Question:

In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 ˚A: (a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from the proton. (b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)? (c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 A˚ separation?

Answer:

(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from the proton.

The potential energy of the system can be estimated using Coulomb’s law:

PE = kq1q2/r

Where k is the Coulomb’s constant, q1 and q2 are the charges of the electron and proton respectively, and r is the distance between them.

In this case, q1 = -1.6 x 10^-19 C, q2 = 1.6 x 10^-19 C, and r = 0.53 x 10^-10 m.

Therefore, PE = 8.99 x 10^9 Nm^2/C^2 x (-1.6 x 10^-19 C) x (1.6 x 10^-19 C) / (0.53 x 10^-10 m)

PE = -2.21 x 10^-18 J

To convert this to eV, we divide by the charge of an electron:

PE = -2.21 x 10^-18 J / (1.6 x 10^-19 C)

PE = -13.8 eV

(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?

The minimum work required to free the electron is equal to the kinetic energy of the electron in its orbit. Since the kinetic energy of the electron is half the magnitude of the potential energy, the minimum work required to free the electron is equal to half the magnitude of the potential energy, or -6.9 eV.

(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 A˚ separation?

(a) The potential energy of the system can be estimated using Coulomb’s law:

PE = kq1q2/r

Where k is the Coulomb’s constant, q1 and q2 are the charges of the electron and proton respectively, and r is the distance between them.

In this case, q1 = -1.6 x 10^-19 C, q2 = 1.6 x 10^-19 C, and r = 1.06 x 10^-10 m.

Therefore, PE = 8.99 x 10^9 Nm^2/C^2 x (-1.6 x 10^-19 C) x (1.6 x 10^-19 C) / (1.06 x 10^-10 m)

PE = -1.08 x 10^-18 J

To convert this to eV, we divide by the charge of an electron:

PE = -1.08 x 10^-18 J / (1.6 x 10^-19 C)

PE = -6.7 eV

(b) The minimum work required to free the electron is equal to the kinetic energy of the electron in its orbit. Since the kinetic energy of the electron is half the magnitude of the potential energy, the minimum work required to free the electron is equal to half the magnitude of the potential energy, or -3.35 eV.

Question:

Answer the following: (a) The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100Vm^−1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage so there is no field inside!) (b) A man fixes outside his house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1m^2. Will he get an electric shock if he touches the metal sheet next morning? (c) The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged? (d) What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning? (Hint: The earth has an electric field of about 100 Vm^−1 at its surface in the downward direction, corresponding to a surface charge density =−10^−9 Cm^−2. Due to the slight conductivity of the atmosphere up to about 50 km (beyond which it is good conductor), about + 1800 C is pumped every second into the earth as a whole. The earth, however, does not get discharged since thunderstorms and lightning occurring continually allover the globe pump an equal amount of negative charge on the earth.)

Answer:

(a) We do not get an electric shock as we step out of our house into the open because the electric field inside the house is zero due to the steel cage. Also, the electric field outside the house is too small to cause an electric shock.

(b) No, the man will not get an electric shock if he touches the metal sheet next morning because the electric field near the surface of the earth is too small to cause an electric shock.

(c) The atmosphere does not discharge itself completely and become electrically neutral because the thunderstorms and lightning occurring continually allover the globe pump an equal amount of negative charge on the earth, which balances out the positive charge that is pumped into the earth every second due to the slight conductivity of the atmosphere.

(d) The electrical energy of the atmosphere is dissipated during a lightning in the form of heat, light, and sound energy.

Question:

A small sphere of radius r1​ and charge q1​ is enclosed by a spherical shell of radius r2​ and charge q2​. Show that if q1​ is positive, charge will necessarily flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q2​ on the shell is.

Answer:

1. Consider the two spheres, one with a radius of r1 and charge q1 and the other with a radius of r2 and charge q2.

2. Assume that q1 is positive and q2 can be either positive or negative.

3. Since q1 is positive, there is an excess of positive charge on the sphere with radius r1.

4. This excess of positive charge will create an electric field that will cause a flow of charge from the sphere with radius r1 to the sphere with radius r2.

5. This flow of charge will continue until the charges on both spheres are equalized.

6. Therefore, no matter what the charge q2 on the shell is, charge will necessarily flow from the sphere to the shell when the two are connected by a wire.

Question:

a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by $$(E_2,-, E_1), \dot, \hat{n},\displaystyle \frac{\sigma }{\varepsilon _0}$$ where n^ is a unit vector normal to the surface at a point and σ is the surface charge density at that point. (The direction of n^ is from side 1 to side 2.) Hence show that just outside a conductor, the electric field is σn^/ϵ0​ (b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another. [Hint: For (a), use Gauss’s law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

Answer:

a) Using Gauss’s law, we can calculate the electric field on either side of the surface.

For side 1,

∮E⃗.dA⃗=Qencl/ϵ₀

E⃗1=Qencl/A₁ϵ₀

For side 2,

∮E⃗.dA⃗=Qencl/ϵ₀

E⃗2=Qencl/A₂ϵ₀

The normal component of the electric field is given by E⃗.n^.

Therefore,

E⃗1.n^=Qencl/A₁ϵ₀.n^

E⃗2.n^=Qencl/A₂ϵ₀.n^

The discontinuity in the normal component of electric field is given by

(E⃗2-E⃗1).n^= (Qencl/A₂ϵ₀.n^) - (Qencl/A₁ϵ₀.n^)

=(Qencl/A₂ϵ₀.n^) - (Qencl/A₁ϵ₀.n^)

=Qencl/A₂ϵ₀.n^ - Qencl/A₁ϵ₀.n^

=(Qencl/A₂ϵ₀ - Qencl/A₁ϵ₀).n^

=(σ/ϵ₀).n^

Hence, the discontinuity in the normal component of electric field from one side of a charged surface to another is given by (E⃗2-E⃗1).n^= (σ/ϵ₀).n^

Just outside a conductor, the electric field is given by

E⃗=σn^/ϵ₀

b) The tangential component of the electric field is given by E⃗xn^.

Therefore,

E⃗1xn^=Qencl/A₁ϵ₀xn^

E⃗2xn^=Qencl/A₂ϵ₀xn^

The discontinuity in the tangential component of electric field is given by

(E⃗2x-E⃗1x).n^= (Qencl/A₂ϵ₀xn^) - (Qencl/A₁ϵ₀xn^)

=(Qencl/A₂ϵ₀xn^) - (Qencl/A₁ϵ₀xn^)

=Qencl/A₂ϵ₀xn^ - Qencl/A₁ϵ₀xn^

=(Qencl/A₂ϵ₀ - Qencl/A₁ϵ₀)xn^

=0

Hence, the discontinuity in the tangential component of electric field from one side of a charged surface to another is zero. Therefore, the tangential component of electrostatic field is continuous from one side of a charged surface to another.

To show this, we can also use the fact that work done by electrostatic field on a closed loop is zero.

Question:

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF =10^−12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Answer:

Step 1: Identify the given values. Given: Capacitance = 8 pF, Dielectric Constant = 6

Step 2: Calculate the new capacitance. New Capacitance = 8 pF x (6/1) = 48 pF

Question:

Two charges 2μC and −2μC are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?

Answer:

a) An equipotential surface is a surface where the electric potential is the same. Therefore, the equipotential surface of the system would be a sphere with a radius of 6 cm, centered at the midpoint between A and B.

b) The direction of the electric field at every point on the equipotential surface would be perpendicular to the surface, since the electric field is always perpendicular to an equipotential surface.

Question:

A 600 pF capacitor is charged by a 200V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?

Answer:

1. Calculate the capacitance of the charged capacitor, C = 600 pF.

2. Calculate the stored energy in the charged capacitor, U = 1/2 x C x V^2 = 1/2 x 600 pF x (200V)^2 = 24,000 J.

3. Calculate the capacitance of the uncharged capacitor, C = 600 pF.

4. Calculate the stored energy in the uncharged capacitor, U = 0 J.

5. Calculate the amount of energy lost in the process by subtracting the stored energy in the uncharged capacitor from the stored energy in the charged capacitor, E = 24,000 J - 0 J = 24,000 J.

Question:

A 4μF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Answer:

1. Calculate the charge stored on the 4μF capacitor: Q = C*V = 4μF * 200V = 800μC

2. Calculate the electrostatic energy stored in the 4μF capacitor: E = ½QV = ½ * 800μC * 200V = 80mJ

3. Calculate the charge stored on the 2μF capacitor after connecting it to the 4μF capacitor: Q = C*V = 2μF * 200V = 400μC

4. Calculate the electrostatic energy stored in the 2μF capacitor: E = ½QV = ½ * 400μC * 200V = 40mJ

5. Calculate the amount of energy lost in the form of heat and electromagnetic radiation: Energy Lost = 80mJ - 40mJ = 40mJ

Question:

If you were a baby crocodile, would you tell Makara that he was wrong? What would you say to convince him?

Answer:

1. First, I would listen carefully to what Makara has to say so that I understand why he believes he is right.

2. Then, I would explain to Makara why I think he is wrong in a respectful way.

3. I would provide evidence to support my argument, such as facts or examples.

4. Finally, I would offer a solution to the problem that Makara is trying to solve, or suggest an alternative point of view.

Question:

Describe schematically the equipotential surfaces corresponding to(a) a constant electric field in the z-direction,(b) a field that uniformly increases in magnitude but remains in a constant (say, z) direction, (c) a single positive charge at the origin, and (d) a uniform grid consisting of long equally spaced parallel charged wires in a plane.

Answer:

(a) The equipotential surfaces will be planes parallel to the z-axis.

(b) The equipotential surfaces will be planes that are increasingly spaced further apart in the z-direction.

(c) The equipotential surfaces will be concentric spheres centered at the origin.

(d) The equipotential surfaces will be parallel planes that are equally spaced apart and perpendicular to the charged wires.

Question:

A regular hexagon of side 10 cm has a charge 5μC at each of its vertices. Calculate the potential at the centre of the hexagon.

Answer:

Step 1: Calculate the total charge at the hexagon. Total charge = 5 μC x 6 vertices = 30 μC

Step 2: Calculate the distance between the centre and each of the vertices. Distance = 10 cm / 2 = 5 cm

Step 3: Calculate the electric field at the centre of the hexagon. Electric field = (30 μC) / (4π x (5 cm)2) = 0.0011 N/C

Step 4: Calculate the potential at the centre of the hexagon. Potential = Electric field x Distance = 0.0011 N/C x 5 cm = 0.0055 V

Question:

Three capacitors each of capacitance 9pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120V supply?

Answer:

(a) Total capacitance of the combination = 9pF x 3 = 27pF

(b) Potential difference across each capacitor = 120V/3 = 40V

Question:

Show that the force on each plate of a parallel plate capacitor has a magnitude equal to (1​/2) QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor.

Answer:

Answer:

The force on each plate of a parallel plate capacitor is the electric force due to the electric field between the plates. The magnitude of the electric field between the plates is equal to E, and the charge on the capacitor is equal to Q.

By Coulomb’s law, the magnitude of the electric force between two charges is equal to (1/4piε)Q1Q2/r^2, where Q1 and Q2 are the charges on the two plates, and r is the distance between the plates.

Since the distance between the plates is constant, the magnitude of the electric force is equal to (1/4piε)Q1Q2/r^2.

Since the charge on each plate is equal to Q, the magnitude of the electric force is equal to (1/4piε)Q^2/r^2.

Since the magnitude of the electric field between the plates is equal to E, the magnitude of the electric force is equal to (1/4piε)Q^2/r^2 = (1/2)QE.

Therefore, the magnitude of the force on each plate of a parallel plate capacitor is equal to (1/2)QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. The origin of the factor 1/2 is due to the fact that the total electric force between the two plates is the sum of the electric forces between the two individual charges.

Question:

What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors are in the range of μF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

Answer:

Step 1: Calculate the area of the plates of the capacitor.

Area = (Separation between the plates) x (Width of the plates)

Step 2: Substitute the given values in the equation.

Area = (0.5 cm) x (Width of the plates)

Step 3: Calculate the area of the plates.

Area = 0.5 cm x (Width of the plates)

Therefore, the area of the plates of the 2 F parallel plate capacitor is 0.5 cm x (Width of the plates).

Step 4: Realise why ordinary capacitors are in the range of μF or less.

The area of the plates of a 2 F parallel plate capacitor is very small (0.5 cm x (Width of the plates)). Therefore, ordinary capacitors are typically in the range of μF or less.

Step 5: Understand why electrolytic capacitors have a much larger capacitance.

Electrolytic capacitors have a much larger capacitance (0.1 F) because of the very minute separation between the conductors.

Question:

A spherical conducting shell of inner radius r1​ and outer radius r2​ has a charge Q. (a) A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell? (b) Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Answer:

a) The surface charge density on the inner surface of the shell is q/4πr1^2 and the surface charge density on the outer surface of the shell is (Q + q)/4πr2^2.

b) No, the electric field inside a cavity with no charge is not necessarily zero, even if the shell is not spherical but has an irregular shape. This is because the electric field inside the cavity is dependent on the shape of the shell, as well as the charge on the shell. If the shell has an irregular shape, it can cause the electric field to be non-zero even if there is no charge inside the cavity.

Question:

A spherical conductor of radius 12 cm has a charge of 1.6×10^−7C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?

Answer:

(a) Inside the sphere, the electric field is zero.

(b) Just outside the sphere, the electric field is equal to the electric field due to a point charge at the centre of the sphere, which can be calculated using Coulomb’s Law: E = kq/r^2, where k is the Coulomb constant (8.99 x 10^9 Nm^2/C^2), q is the charge on the sphere (1.6 x 10^-7 C), and r is the radius of the sphere (12 cm). Therefore, the electric field just outside the sphere is E = 8.99 x 10^9 Nm^2/C^2 x 1.6 x 10^-7 C / (12 cm)^2 = 7.495 x 10^3 N/C.

(c) At a point 18 cm from the centre of the sphere, the electric field is again equal to the electric field due to a point charge at the centre of the sphere, which can be calculated using Coulomb’s Law: E = kq/r^2, where k is the Coulomb constant (8.99 x 10^9 Nm^2/C^2), q is the charge on the sphere (1.6 x 10^-7 C), and r is the distance from the centre of the sphere (18 cm). Therefore, the electric field at a point 18 cm from the centre of the sphere is E = 8.99 x 10^9 Nm^2/C^2 x 1.6 x 10^-7 C / (18 cm)^2 = 4.996 x 10^3 N/C.

Question:

Was the old woman’s gift to Vijay Singh eccentric? Why?

Answer:

Step 1: Identify the facts of the question. Facts: An old woman gave a gift to Vijay Singh.

Step 2: Determine what is meant by “eccentric”. Eccentric: Unusual or strange in behavior, appearance, or ideas.

Step 3: Determine if the gift was eccentric. Answer: It is not possible to determine if the old woman’s gift to Vijay Singh was eccentric without additional information.

Question:

Two charges 5×10−8 C and −3×10^−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Answer:

1. Calculate the electric field due to each charge using Coulomb’s law: E1 = (k510^-8)/(16^2) E2 = (k*-3*10^-8)/(16^2)

2. Calculate the total electric field: E = E1 + E2

3. Set the total electric field equal to zero: E = 0

4. Solve for the x-coordinate of the point: x = (k810^-8)/(E1 - E2)

5. The point at which the electric potential is zero is located at x.

Question:

Exercise: In a parallel plate capacitor with air between the plates, each plate has an area of 6×10^−3m^2 and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor? Explain what would happen if in the capacitor given in Exercise a 3mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) While the voltage supply remained connected. (b) After the supply was disconnected.

Answer:

Exercise:

Capacitance of the capacitor = ε0 * A/d = 8.85 * 10^-12 * 6 * 10^-3 / 0.003 = 35.4 * 10^-9 F

Charge on each plate of the capacitor = capacitance * voltage = 35.4 * 10^-9 * 100 = 3.54 * 10^-7 C

(a) If a 3mm thick mica sheet (of dielectric constant = 6) were inserted between the plates while the voltage supply remained connected, the capacitance of the capacitor would increase. The capacitance of the capacitor can be calculated as follows:

Capacitance of the capacitor = εr * ε0 * A/d = 6 * 8.85 * 10^-12 * 6 * 10^-3 / 0.003 = 211.4 * 10^-9 F

The charge on each plate of the capacitor would also increase to:

Charge on each plate of the capacitor = capacitance * voltage = 211.4 * 10^-9 * 100 = 2.114 * 10^-6 C

(b) If the supply was disconnected after the mica sheet was inserted, the capacitance of the capacitor would remain the same (211.4 * 10^-9 F). However, the charge on each plate of the capacitor would be zero as the voltage supply is disconnected.

Question:

An electrical technician requires a capacitance of 2μF in a circuit across a potential difference of 1 kV. A large number of 1μF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.

Answer:

1. Calculate the required capacitance in terms of the available capacitors: 2μF = 2 x 1μF

2. Determine the minimum number of capacitors required: 2 x 1μF = 2 capacitors

3. Calculate the total potential difference that the capacitors must withstand: 1 kV = 1000 V

4. Determine the maximum potential difference that each capacitor can withstand: 400 V

5. Suggest a possible arrangement: The two 1μF capacitors should be connected in series, so that the total capacitance is 2μF and the total potential difference is 800 V, which is within the maximum potential difference of each capacitor.

Question:

A long charged cylinder of linear charged density λ is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?

Answer:

1. Identify the physical parameters of the problem, including the linear charge density λ, the inner and outer radii of the two cylinders, and the distance between them.

2. Calculate the total charge on the inner cylinder, using the equation Q = λπr^2, where r is the radius of the inner cylinder.

3. Use Gauss’s Law to calculate the electric field in the space between the two cylinders. The electric field is given by the equation E = λ/2πεo, where εo is the permittivity of free space.

4. Calculate the electric field in the space between the two cylinders, using the equation E = Q/Aεo, where Q is the total charge on the inner cylinder and A is the area of the space between the two cylinders.

Question:

Two tiny spheres carrying charges 1.5μC and 2.5μC are located 30 cm apart. Find the potential and electric field:

Answer:

Answer:

1. Calculate the electric field: E = kq1q2/r^2 E = 8.9910^9 Nm^2/C^2 * 1.510^-6 C * 2.510^-6 C / (0.3 m)^2 E = 6.6610^5 N/C

2. Calculate the potential: V = kq1/r + kq2/r V = 8.9910^9 Nm^2/C^2 * (1.510^-6 C / 0.3 m) + 8.9910^9 Nm^2/C^2 * (2.510^-6 C / 0.3 m) V = 8.33*10^5 V

Question:

(a) at the mid-point of the line joining the two charges, and (b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Answer:

A.

1. Calculate the distance between the two charges.
2. Calculate the mid-point of the line joining the two charges.
3. Calculate the point 10 cm from the mid-point in a plane normal to the line and passing through the mid-point.

Question:

Two charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.

Answer:

1. The ratio of electric fields at the surfaces of two spheres is equal to the inverse of the ratio of their radii, i.e. Ea/Eb = b/a.

2. This result can be used to explain why charge density on sharp and pointed ends of a conductor is higher than on its flatter portions. This is because the electric field at the surface of a sharp and pointed end is greater than that on its flatter portions due to its smaller radius. Therefore, the charge density at the sharp and pointed end is higher than on its flatter portions.

Question:

If one of the two electrons of a H2​ molecule is removed, we get a hydrogen molecular ion H2^+​. In the ground state of an H2^+​, the two protons are separated by roughly 1.5A˚ , and the electron is roughly 1A˚ from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.

Answer:

Answer:

1. Calculate the Coulombic potential energy of the system:

PE = (k * q1 * q2) / r

where k = 8.99 x 10^9 N*m^2/C^2 q1 and q2 are the charges of the protons in the H2+ molecule (both have a charge of +1) r is the distance between the protons (1.5 A˚)

PE = (8.99 x 10^9 N*m^2/C^2) * (1 C)^2 / (1.5 A˚)

PE = 5.99 x 10^9 Joules

2. Specify the zero of potential energy:

The zero of potential energy is the energy of the two separated protons and the electron in the H2+ molecule when the protons and electron are infinitely far apart.

Question:

A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Answer:

Step 1: Calculate the total charge of the cube. Total charge = q x 8 (vertices) = 8q

Step 2: Calculate the potential at the centre of the cube due to the charge array. Potential at centre of cube = (k x 8q)/b (where k is the Coulomb constant)

Step 3: Calculate the electric field at the centre of the cube due to the charge array. Electric field at centre of cube = (k x 8q)/b2 (where b is the side of the cube)

Question:

Answer carefully: (a) Two large conducting spheres carrying charges Q1​ and Q2​ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by Q1​Q2​/4πϵ0​r^2, where r is the distance between their centres? (b) If Coulomb’s law involved 1/r^3 dependence (instead of 1/r^2),would Gauss’ law be still true ? (c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point? (d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical? (e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there? (f) What meaning would you give to the capacitance of a single conductor? (g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

Answer:

(a) Yes, the magnitude of electrostatic force between them is exactly given by Q1Q2/4πϵ0r^2, where r is the distance between their centres.

(b) No, if Coulomb’s law involved 1/r^3 dependence, Gauss’ law would not be true.

(c) Yes, the test charge will travel along the field line passing through that point.

(d) The work done by the field of a nucleus in a complete circular orbit of the electron is zero. If the orbit is elliptical, the work done will depend on the shape and size of the ellipse.

(e) Yes, electric potential is also discontinuous across the surface of a charged conductor.

(f) The capacitance of a single conductor is zero since it cannot store any charge.

(g) Water has a much greater dielectric constant than mica because it is a polar molecule, meaning that it has a positive charge at one end and a negative charge at the other end. This allows it to absorb and store more electric charge than non-polar molecules such as mica.

Question:

A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 107 Vm^−1. (Dielectric strength is the maximum electric field a material can tolerate without breakdown, i.e., without starting to conduct electricity through partial ionisation.) For safety, we should like the field never to exceed, say 10% of the dielectric strength.What minimum area of the plates is required to have a capacitance of 50 pF ?

Answer:

1. Calculate the maximum electric field: Maximum electric field = 0.1 x 107 Vm^−1 = 10.7 Vm^−1

2. Calculate the electric field between the plates: Electric field between the plates = 1 kV/50 pF = 20,000 Vm^−1

3. Calculate the minimum area of the plates: Minimum area of the plates = Electric field between the plates/Maximum electric field = 20,000 Vm^−1/10.7 Vm^−1 = 1,867 m^2

Question:

A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5μC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends).

Answer:

1. Calculate the area of the inner cylinder: A = πr2 = 3.14 x 1.42 = 6.09 cm2

2. Calculate the area of the outer cylinder: A = πr2 = 3.14 x 1.52 = 7.07 cm2

3. Calculate the gap between the two cylinders: d = (15 cm - (2 x 1.5 cm)) = 12 cm

4. Calculate the capacitance of the system: C = ε0εrA/d = 8.85 x 10-12 x 7.07/12 = 5.52 x 10-12 F

5. Calculate the potential of the inner cylinder: V = Q/C = 3.5 x 10-6/5.52 x 10-12 = 6.3 x 106 V

Question:

A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5μC. The space between the concentric spheres is filled with a liquid of dielectric constant 32. (a) Determine the capacitance of the capacitor. (b) What is the potential of the inner sphere? (c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.

Answer:

(a) The capacitance of the spherical capacitor can be calculated using the formula C = 4πεεo r, where ε is the dielectric constant of the liquid and r is the radius of the inner sphere. Substituting the given values, the capacitance of the capacitor is C = 4π(32)(8.85 x 10^-12)(12 x 10^-2) = 0.0089 F.

(b) The potential of the inner sphere can be calculated using the formula V = Q/C, where Q is the charge and C is the capacitance. Substituting the given values, the potential of the inner sphere is V = (2.5 x 10^-6)/(0.0089) = 2.8 V.

(c) The capacitance of an isolated sphere of radius 12 cm is much smaller than the capacitance of the spherical capacitor because the capacitance of a capacitor is dependent on the dielectric constant of the material between the two charges. The dielectric constant of the liquid in the spherical capacitor is much higher than that of an isolated sphere, resulting in a much higher capacitance.

Question:

A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Answer:

Step 1: Calculate the capacitance (C) of the capacitor, which is equal to 12 pF.

Step 2: Calculate the voltage (V) of the battery, which is equal to 50 V.

Step 3: Calculate the electrostatic energy (E) stored in the capacitor using the equation E = 1/2CV2, which is equal to 1/2 x 12 pF x 502 = 1500 pFV2.

Step 4: Convert the energy to joules by multiplying the result by 1 pFV2 = 1.113×10-12 J, which is equal to 1.67×10-9 J.

Therefore, the electrostatic energy stored in the capacitor is 1.67×10-9 J.

Question:

The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply. (a) How much electrostatic energy is stored by the capacitor? (b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates

Answer:

(a) Electrostatic energy stored by the capacitor = ½ × capacitance × voltage²

Capacitance of the capacitor = ε × area/separation

= 8.854 × 10⁻¹² × 90/2.5 × 10⁻³

= 3.543 × 10⁻⁸ F

Therefore, electrostatic energy stored by the capacitor = ½ × 3.543 × 10⁻⁸ × 400²

= 0.014 J

(b) Energy per unit volume u = electrostatic energy/volume of capacitor

Volume of capacitor = area × separation

= 90 × 2.5 × 10⁻³

= 0.225 cm³

Therefore, energy per unit volume u = 0.014/0.225

= 0.062 J/cm³

Relation between u and electric field E = u = ½ ε E²

Question:

Two charges -q and +q are located at points (0, 0, -a) and (0, 0, a), respectively. (a) What is the electrostatic potential at the points (0, 0, z) and (x, y, 0) ? (b) Obtain the dependence of potential on the distance r of a point from the origin when r/a » 1 (c) How much work is done in moving a small test charge from the point (5,0,0) to (-7,0,0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?

Answer:

(a) The electrostatic potential at the points (0, 0, z) and (x, y, 0) can be calculated using the equation V = k * (q1/r1 + q2/r2). Here, k is the Coulomb’s constant, q1 and q2 are the charges of the two charges at (0, 0, -a) and (0, 0, a) respectively, and r1 and r2 are the distances from the charges to the points (0, 0, z) and (x, y, 0).

(b) The potential at a point at a distance r from the origin can be obtained by substituting r for r1 and r2 in the equation V = k * (q1/r1 + q2/r2). When r/a » 1, the potential can be approximated as V = 2kq/r.

(c) The work done in moving a small test charge from the point (5,0,0) to (-7,0,0) along the x-axis can be calculated using the equation W = ∫Fdx. Here, F is the electrostatic force between the two charges and x is the distance between the two points. If the path of the test charge between the same points is not along the x-axis, the work done will be different since the distance between the two points will be different.

Question:

Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Answer:

(a) Total Capacitance = 2 pF + 3 pF + 4 pF = 9 pF

(b) Charge on each capacitor = Total Charge/Number of Capacitors

Charge on each capacitor = 100 V/9 pF = 11.11 V