Question:

Two point charges qA​=3μC and qB​=−3μC are located 20 cm apart in vacuum (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude 1.5×10^−9C is placed at this point, what is the force experienced by the test charge?

Answer:

a) The electric field at the midpoint O of the line AB joining the two charges is zero. This is because the electric field due to qA​ and qB​ cancel each other out at the midpoint.

b) The force experienced by the test charge is zero, because the electric field at the midpoint is zero.

Question:

Check that the ratio ke^2 /G me​mp​ is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?

Answer:

1. First, check that the ratio ke^2/Gmep is dimensionless. To do this, calculate the dimensions of each term in the ratio. ke^2 has dimensions of [ML^2T^-2], G has dimensions of [ML^3T^-2], and mep has dimensions of [ML^2T^-2]. Since the dimensions of the two terms in the ratio are the same, the ratio is dimensionless.

2. Look up a Table of Physical Constants and determine the value of this ratio. The value of the ratio ke^2/Gmep is approximately 8.98755 x 10^9.

3. This ratio signifies the electrostatic force between two protons. The value of the ratio is equal to the electrostatic force (in Newtons) between two protons that are separated by a distance of 1 meter.

Question:

Four point charges qA​=2μC,qB​=−5μC,qC​=2μC, and qD​=−5μCare located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1μC placed at the centre of the square?

Answer:

1. Calculate the distance between each of the four charges and the centre of the square: A: 5 cm B: 7.07 cm C: 5 cm D: 7.07 cm

2. Use Coulomb’s law to calculate the force between each charge and the centre of the square: A: F_A = (8.99 x 10^9 Nm^2/C^2) x (2 x 10^-6 C) / (5 x 10^-2 m)^2 = 7.192 x 10^-5 N B: F_B = (8.99 x 10^9 Nm^2/C^2) x (-5 x 10^-6 C) / (7.07 x 10^-2 m)^2 = -3.938 x 10^-5 N C: F_C = (8.99 x 10^9 Nm^2/C^2) x (2 x 10^-6 C) / (5 x 10^-2 m)^2 = 7.192 x 10^-5 N D: F_D = (8.99 x 10^9 Nm^2/C^2) x (-5 x 10^-6 C) / (7.07 x 10^-2 m)^2 = -3.938 x 10^-5 N

3. Add the forces together to calculate the total force on the charge of 1μC placed at the centre of the square: F_total = F_A + F_B + F_C + F_D = 7.192 x 10^-5 N + (-3.938 x 10^-5 N) + (7.192 x 10^-5 N) + (-3.938 x 10^-5 N) = 0

Therefore, the force on a charge of 1μC placed at the centre of the square is 0 N.

Question:

a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable. (b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.

Answer:

a) At a null point, the electric field is zero. This means that there is no net force acting on the test charge. If the test charge is moved from its equilibrium position, the electric field will no longer be zero, and a net force will be exerted on the test charge. This net force will act to move the test charge away from its equilibrium position, thus making the equilibrium of the test charge unstable.

b) To verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart, we can calculate the electric field at the null point. The electric field at a point due to a single point charge is given by E = kq/r^2, where k is the Coulomb constant, q is the charge of the point charge, and r is the distance from the point charge to the point at which the electric field is being calculated. The electric field at the null point is the sum of the electric fields due to each charge. Since the charges have the same magnitude and sign, they will cancel each other out, resulting in a net electric field of zero. This confirms that the equilibrium of the test charge is unstable, as the electric field at the null point is zero.

Question:

(a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5×10^−7 C? The radii of A and B are negligible compared to the distance of separation. (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

Answer:

(a) The force of electrostatic repulsion between two charged spheres is given by Coulomb’s Law:

F = K * (Q1 * Q2) / (r^2)

Where K is Coulomb’s constant (8.99 x 10^9 N m^2/C^2), Q1 and Q2 are the charges on the two spheres, and r is the distance between them.

Substituting the given values, we get:

F = 8.99 x 10^9 N m^2/C^2 * (6.5 x 10^-7 C)^2 / (50 cm)^2

F = 2.71 x 10^-4 N

(b) For the second part, we need to double the charge and halve the distance.

F = 8.99 x 10^9 N m^2/C^2 * (2 * 6.5 x 10^-7 C)^2 / (25 cm)^2

F = 10.85 x 10^-4 N

Question:

Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0×10^3Nm^2/C.(a) What is the net charge inside the box?(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?

Answer:

(a) The net charge inside the box can be calculated using the equation: net outward flux = ε₀E₀A, where ε₀ is the permittivity of free space, E₀ is the electric field, and A is the area of the box. Rearranging the equation, we get: net charge = (8.0×10^3Nm^2/C)/(ε₀E₀A).

(b) No, you cannot conclude that there are no charges inside the box if the net outward flux is zero. This is because the electric field can be zero even if there are charges inside the box.

Question:

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55×10^4NC^−1 in Millikans oil drop experiment. The density of the oil is 1.26 g cm^−3. Estimate the radius of the drop. [g = 9.8 ms^−2; e=1.60×10^−19 C].

Answer:

1. Calculate the charge of the oil drop using the equation e = charge of the oil drop: e = 12 x 1.60 x 10^-19 C = 1.92 x 10^-18 C

2. Calculate the electric field strength using the equation E = F/q: E = 2.55 x 10^4 N/C / 1.92 x 10^-18 C = 1.32 x 10^22 N/C

3. Calculate the force acting on the oil drop using the equation F = qE: F = 1.92 x 10^-18 C x 1.32 x 10^22 N/C = 2.53 x 10^4 N

4. Calculate the mass of the oil drop using the equation m = ρV: m = 1.26 g/cm^3 x (4/3)πr^3 = 4/3πr^3

5. Calculate the radius of the oil drop using the equation F = ma: F = ma 2.53 x 10^4 N = 4/3πr^3 x 9.8 m/s^2 r = (2.53 x 10^4 N / (4/3π x 9.8 m/s^2))^1/3 r = 1.59 x 10^-8 m

Question:

It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so-called ‘up’ quark (denoted by u) of charge + (2/3) e, and the ‘down’ quark (denoted by d) of charge (-1/3) e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.

Answer:

1. A proton consists of two ‘up’ quarks (denoted by u) and one ‘down’ quark (denoted by d).
2. A neutron consists of one ‘up’ quark (denoted by u) and two ‘down’ quarks (denoted by d).
3. Therefore, the quark composition of a proton is uud and the quark composition of a neutron is udd.

Question:

What is the force between two small charged spheres having charges of 2×10^−7C and 3×10^−7C placed 30 cm apart in air?

Answer:

Step 1: Calculate the electric field (E) between the two charged spheres. This can be done using Coulomb’s Law, which states that the force between two charged particles is equal to the product of the charges (q1 and q2) divided by the distance (r) between them, squared:

E = (q1q2)/(4piepsilonr^2)

Step 2: Calculate the force (F) between the two charged spheres. This can be done using Newton’s Law of Universal Gravitation, which states that the force between two objects is equal to the product of their masses (m1 and m2) divided by the distance between them, squared:

F = (m1*m2)/(r^2)

Step 3: Substitute the values for q1, q2, r, and epsilon into the equation for E to calculate the electric field between the two charged spheres:

E = (2×10^−7C * 3×10^−7C)/(4pi8.85e-12*(30 cm)^2)

Step 4: Substitute the values for m1, m2, and r into the equation for F to calculate the force between the two charged spheres:

F = (2×10^−7C * 3×10^−7C)/((30 cm)^2)

Step 5: The force between the two charged spheres is equal to 1.5×10^−7N.

Question:

The electrostatic force on a small sphere of charge 0.4μC due to another small sphere of charge −0.8μC in air is 0.2 N. (a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?

Answer:

a) The distance between the two spheres can be determined using Coulomb’s Law: F = k*(q1q2)/(r^2) where F is the electrostatic force, k is the Coulomb’s constant (8.9910^9 N*m^2/C^2), q1 and q2 are the charges of the spheres, and r is the distance between the two spheres.

Substituting the given values, we get: 0.2 = 8.9910^9(0.4*(-0.8))/r^2

Solving for r, we get: r = 0.4 m

b) The force on the second sphere due to the first can be determined using Coulomb’s Law: F = k*(q1*q2)/(r^2)

Substituting the given values, we get: F = 8.9910^9(0.4*(-0.8))/(0.4^2)

Solving for F, we get: F = -1.8 N

Question:

Consider a uniform electric field E=3×10^3 i^ NC^−1. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60o angle with the x-axis?

Answer:

a) The flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane is 30 NC.

b) The flux through the same square if the normal to its plane makes a 60o angle with the x-axis is 15 NC.

Question:

A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5×10^3 N/C and points radially inward, what is the net charge on the sphere?

Answer:

1. Calculate the electric field at the surface of the sphere using Coulomb’s law: E = (k*Q)/(r^2)

2. Solve for the charge on the sphere: Q = (E*r^2)/k Q = (1.5x10^3 N/C * (0.1 m)^2)/(8.99x10^9 N m^2/C^2) Q = 1.67x10^-7 C

Question:

A system has two charges qA​=2.5×10^−7C and qB​=−2.5×10^−7C located at points A:(0,0,15cm) and B:(0,0,+15cm), respectively. What are the total charge and electric dipole moment of the system?

Answer:

Total charge: qT = qA + qB = 2.5 × 10^−7C + (-2.5 × 10^−7C) = 0C

Electric dipole moment: p = qA * rA - qB * rB = (2.5 × 10^−7C) * (0, 0, 15cm) - (-2.5 × 10^−7C) * (0, 0, +15cm) = (0, 0, 30 × 10^−7C m)

Question:

Exercise:[Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. The charge on each is 6.5×10^7 C? The radii of A and B are negligible compared to the distance of separation.] Suppose the spheres A and B in Exercise have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?

Answer:

Answer:

Step 1: Calculate the original force of repulsion between the two spheres A and B:

F = (kq1q2)/r^2 = (910^9 * 6.510^7 * 6.510^7)/(5010^-2)^2 = 2.1*10^7 N

Step 2: Calculate the charge on the third sphere:

q3 = q1 + q2 = 26.510^7 C

Step 3: Calculate the new force of repulsion between the two spheres A and B:

F’ = (kq1q3)/r^2 = (910^9 * 6.510^7 * 26.510^7)/(5010^-2)^2 = 4.210^7 N

Therefore, the new force of repulsion between A and B is 4.2*10^7 N.

Question:

Consider a uniform electric field E=3×10^3 i^ NC−1. What is the net flux of the uniform electric field through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

Answer:

Answer:

Step 1: Calculate the area of one face of the cube: A = (20 cm)^2 = 400 cm^2

Step 2: Calculate the total area of the cube: Total area = 6 x A = 6 x 400 cm^2 = 2400 cm^2

Step 3: Calculate the net flux of the uniform electric field: Flux = E x Total area = 3 x 10^3 x 2400 cm^2 = 7.2 x 10^6 NC

Question:

A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/2ϵ0​)n^ where n^ is the unit vector in the outward normal direction, and σ is the surface charge density near the hole.

Answer:

1. Begin by recalling Gauss’ Law for electric fields, which states that the flux of the electric field through any closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space (ϵ0).

2. Consider a small surface area element dA at the hole in the conductor. The electric field in the hole is given by the flux of the electric field through this surface element, which can be written as E⃗⋅dA⃗=σdA/ϵ0.

3. Since the unit vector n^ is in the outward normal direction, the electric field in the hole can be written as E⃗⋅n^dA=σdA/ϵ0.

4. Rearranging this equation gives E⃗⋅n^=σ/ϵ0.

5. Finally, substituting in the unit vector n^ gives the result that the electric field in the hole is (σ/2ϵ0​)n^.

Question:

When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Answer:

1. The law of conservation of charge states that the net charge of an isolated system remains constant over time.

2. When a glass rod is rubbed with a silk cloth, electrons are transferred from the glass rod to the silk cloth, creating a negative charge on the glass rod and a positive charge on the silk cloth.

3. This transfer of electrons is consistent with the law of conservation of charge, as the total charge of the system remains the same (i.e. the net charge of the system is conserved).

4. The same phenomenon is observed with many other pairs of bodies, as the transfer of electrons is consistent with the law of conservation of charge.

Question:

An infinite line charge produces a field of 9×10^4 N/C at a distance of 2 cm. Calculate the linear charge density.

Answer:

Step 1: Identify the given variables. Given: Electric field (E) = 9×10^4 N/C, Distance (d) = 2 cm

Step 2: Use the formula for linear charge density. Linear charge density (λ) = E/d

Step 3: Substitute the given values in the formula. λ = (9×10^4 N/C)/(2 cm)

Step 4: Calculate the linear charge density. λ = 4.5×10^4 C/m

Question:

(a) Explain the meaning of the statement ’electric charge of a body is quantised’. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?

Answer:

(a) The statement ’electric charge of a body is quantised’ means that electric charge can only exist in discrete, indivisible amounts, rather than in a continuous range of values.

(b) Quantisation of electric charge can be ignored when dealing with macroscopic charges because the effects of quantisation are only significant on the atomic and subatomic scale. On the macroscopic scale, the discrete nature of electric charge is not relevant, and the charge can be assumed to be continuous.

Question:

An electric dipole with dipole moment 4×10^−9C m is aligned at 30∘ with the direction of a uniform electric field of magnitude 5×10^4NC^−1. Calculate the magnitude of the torque acting on the dipole.

Answer:

1. Convert the dipole moment to SI units: 4 × 10^−9 C m = 4 × 10^−9 N m

2. Calculate the magnitude of the electric field: 5 × 10^4 NC^−1 = 5 × 10^8 N C^−1

3. Calculate the magnitude of the force acting on the dipole: F = 4 × 10^−9 N m × 5 × 10^8 N C^−1 = 2 × 10^−1 N

4. Calculate the magnitude of the torque acting on the dipole: τ = F × sin(30°) = 2 × 10^−1 N × 0.5 = 1 × 10^−1 N m

Question:

A point charge of 2.0μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

Answer:

1. Calculate the electric field at the centre of the cube using Coulomb’s law: E = kQ/r2, where k is the Coulomb constant, Q is the charge, and r is the distance from the centre of the cube to the point charge.

2. Calculate the area of each face of the cube using the length of the edge: A = l2, where l is the length of the edge.

3. Calculate the electric flux through each face of the cube using the equation Φ = EA, where E is the electric field and A is the area of the face.

4. Add up the electric flux of each face to get the total electric flux through the cube: Φtotal = Φ1 + Φ2 + Φ3 + Φ4 + Φ5 + Φ6.

5. The net electric flux through the cube is the total electric flux minus the electric flux through the opposite faces. For example, the electric flux through the top face minus the electric flux through the bottom face.

Question:

Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law.[Hint: Use Coulombs law directly and evaluate the necessary integral.]

Answer:

1. Recall that the electric field due to a point charge q is given by Coulomb’s law: E = kq/r2, where k is the Coulomb’s constant and r is the distance from the point charge.

2. Since the wire has a uniform linear charge density, the total charge on the wire is given by the integral of the linear charge density over the length of the wire.

3. Thus, the electric field due to the wire is given by the integral of the electric field due to each point charge along the wire, multiplied by the linear charge density: E = kλ∫1/r2dr, where r is the distance from the point charge.

4. Evaluate the integral to obtain the formula for the electric field due to the wire: E = kλ/r

Question:

In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 10^5NC^−1 m^−1 per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10^−7 Cm in the negative z-direction ?

Answer:

1. Calculate the magnitude of the total dipole moment: 10^−7 Cm

2. Determine the magnitude of the electric field in the region: 10^5 NC^−1 m^−1 per metre

3. Calculate the force experienced by the system: 10^−7 Cm x 10^5 NC^−1 m^−1 = 10^−2 N

4. Calculate the torque experienced by the system: 10^−7 Cm x 10^5 NC^−1 m^−1 x 1 m = 10^−2 Nm

Question:

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0×10^−22 C/m^2. What is E: (a) in the outer region of the first plate. (b) in the outer region of the second plate, and (c) between the plates?

Answer:

a) In the outer region of the first plate, E = 0 because the electric field is only present between the two plates.

b) In the outer region of the second plate, E = 0 because the electric field is only present between the two plates.

c) Between the plates, E = (17.0×10^−22 C/m^2) / (8.85 * 10^-12 C^2/Nm^2) = 1.9 * 10^9 N/C

Question:

A point charge causes an electric flux of −1.0×10^3 Nm^2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled,how much flux would pass through the surface? (b) What is the value of the point charge?

Answer:

a) If the radius of the Gaussian surface were doubled, 4×10^3 Nm^2/C of flux would pass through the surface.

b) The value of the point charge is not given, and cannot be determined from the given information.

Question:

A polythene piece rubbed with wool is found to have a negative charge of 3×10^−7C. a) Estimate the number of electrons transferred (from which to which?) b) Is there a transfer of mass from wool to polythene?

Answer:

a) Estimate the number of electrons transferred: The charge of a single electron is 1.6 x 10^-19 C, so the number of electrons transferred is 3 x 10^-7 C / 1.6 x 10^-19 C = 1.875 x 10^12 electrons.

b) Is there a transfer of mass from wool to polythene? No, there is no transfer of mass from wool to polythene. The charge transfer is due to the transfer of electrons from one material to the other, not the transfer of mass.

Question:

A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80 μC/m^2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

Answer:

a) Charge on the sphere = πr^2 x surface charge density = 3.14 x (1.2 m)^2 x 80 μC/m^2 = 4,259.2 μC

b) Total electric flux leaving the surface of the sphere = surface charge density x surface area = 80 μC/m^2 x 4πr^2 = 4π x (1.2 m)^2 x 80 μC/m^2 = 75,398.4 μC m^2