Electrostatic Potential And Potential Energy - Part 3

• Recap:
  • Electrostatic potential
  • Electric potential energy
• Today’s topic: Calculation of potential energy using the electrostatic potential
• Investigating a point charge
  • Consider a point charge q at position P in the electric field created by another charge Q.
  • The potential energy of the charge q can be calculated using the formula:
    • Electric potential energy (U) = qV, where V is the electrostatic potential at point P
  • Let’s solve some problems to understand this concept better

Example 1

• Problem:

  • A charge of +3 μC is placed 1 m away from a point charge of +5 μC. Calculate the potential energy of the system.

• Solution:

  • Given:
    • Charge of q1 = +3 μC, charge of q2 = +5 μC
    • Distance between charges, r = 1 m
  • The electrostatic potential at point P due to q2 is given by:
    • V = k * q2 / r (where k is the electrostatic constant, approximately 9 × 10^9 N m^2 C^-2)
  • Calculating the potential energy:
    • U = q1 * V
    • U = (3 × 10^-6 C) * (9 × 10^9 N m^2 C^-2) * (5 × 10^-6 C) / (1 m)
    • U = 135 N m

Example 2

• Problem:

  • Two point charges +2 μC and -4 μC are placed 0.5 m apart. Calculate the potential energy of the system.

• Solution:

  • Given:
    • Charge of q1 = +2 μC, charge of q2 = -4 μC
    • Distance between charges, r = 0.5 m
  • The electrostatic potential at point P due to q2 is given by:
    • V = k * q2 / r
  • Calculating the potential energy:
    • U = q1 * V
    • U = (2 × 10^-6 C) * (9 × 10^9 N m^2 C^-2) * (-4 × 10^-6 C) / (0.5 m)
    • U = -144 N m

Electric Potential Difference

• The concept of electric potential difference is important when discussing potential energy.
• Electric potential difference is the change in electric potential energy per unit charge.
• It is denoted by ∆V and given by the equation:
  • ∆V = Vf - Vi (where Vf is the final potential and Vi is the initial potential)

Potential Difference due to a Point Charge

• The potential difference due to a point charge can be calculated using the formula:
  • ∆V = k * (q / r2 - q / r1) (where q is the charge, r1 is the initial distance, and r2 is the final distance)

Example 3

• Problem:

  • A charge of +4 μC is moved from a distance of 2 m to a distance of 1 m from a point charge of +2 μC. Calculate the potential difference.

• Solution:

  • Given:
    • Charge of q1 = +4 μC, charge of q2 = +2 μC
    • Initial distance, r1 = 2 m, final distance, r2 = 1 m
  • Calculating the potential difference:
    • ∆V = k * (q2 / r2 - q2 / r1)
    • ∆V = (9 × 10^9 N m^2 C^-2) * ((2 × 10^-6 C) / (1 m) - (2 × 10^-6 C) / (2 m))
    • ∆V = 9 × 10^9 N m^2 C^-2

Example 4

• Problem:

  • A charge of +10 μC is moved from a distance of 1 m to a distance of 3 m from a point charge of -4 μC. Calculate the potential difference.

• Solution:

  • Given:
    • Charge of q1 = +10 μC, charge of q2 = -4 μC
    • Initial distance, r1 = 1 m, final distance, r2 = 3 m
  • Calculating the potential difference:
    • ∆V = k * (q2 / r2 - q2 / r1)
    • ∆V = (9 × 10^9 N m^2 C^-2) * ((-4 × 10^-6 C) / (3 m) - (-4 × 10^-6 C) / (1 m))
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Potential Energy due to a System of Charges

• When multiple charges are present in a system, the total potential energy of the system is the sum of the potential energies of each individual charge.
• The potential energy of a system of charges is given by the equation:
  • U = Σ (qi * V), where qi is the charge at each point and V is the electrostatic potential at that point.

Example 5

• Problem:

  • Consider three point charges: +2 μC, -3 μC, and +5 μC. Calculate the total potential energy of the system.

• Solution:

  • Given charges: q1 = +2 μC, q2 = -3 μC, q3 = +5 μC
  • Potential at each point:
    • V1 = k * (q2 / r12 + q3 / r13)
    • V2 = k * (q1 / r21 + q3 / r23)
    • V3 = k * (q1 / r31 + q2 / r32)
  • Total potential energy:
    • U = q1 * V1 + q2 * V2 + q3 * V3

Equipotential Surfaces

• An equipotential surface is a surface on which all points have the same electric potential.
• The electric field lines are always perpendicular to the equipotential surfaces.
• The work done in moving a charge along an equipotential surface is zero, as the potential remains constant.
• Equipotential surfaces are represented by parallel lines or equipotential contours in diagrams.

Calculation of Potential on an Equipotential Surface

• The potential on an equipotential surface can be calculated using the formula:
  • V = k * (q1 / r1 + q2 / r2 + q3 / r3 + …) (where q1, q2, q3, … are the charges and r1, r2, r3, … are the distances from each charge to the point of interest)

Example 6

• Problem:

  • Two charges, +6 μC and -2 μC, are placed at distances of 2 m and 4 m, respectively, from a point in space. Calculate the potential on the resulting equipotential surface.

• Solution:

  • Given charges: q1 = +6 μC, q2 = -2 μC
  • Distances: r1 = 2 m, r2 = 4 m
  • Calculating the potential:
    • V = k * (q1 / r1 + q2 / r2)
    • V = (9 × 10^9 N m^2 C^-2) * ((6 × 10^-6 C) / (2 m) + (-2 × 10^-6 C) / (4 m))

Electric Potential due to a Continuous Charge Distribution

• In some cases, charges may be distributed continuously along a line, surface, or volume.
• To calculate the electric potential due to such continuous charge distributions, the concept of integration is used.
• The electric potential at a point due to a continuous charge distribution is given by the equation:
  • V = k ∫(dq / r) (where dq represents an infinitesimally small charge element and the integral is taken over the entire distribution)

Example 7

• Problem:

  • A uniformly charged rod of length L has a total charge Q. Calculate the electric potential at a point P located at a distance x from one end of the rod.

• Solution:

  • Given:
    • Total charge of the rod, Q
    • Distance from one end, x
    • Length of the rod, L
  • The electric potential at point P can be calculated using the equation:
    • V = k ∫(dq / r)
  • Integrating over the entire rod:
    • V = k ∫(Q / L) / (x^2 + r^2)^0.5
    • V = k * (Q / L) * ln((L + (x^2 + L^2)^0.5) / x)

Potential due to a Charged Disk

• A charged disk is another example of a continuous charge distribution.
• To calculate the electric potential at a point outside the disk, the concept of integration is again used.
• The electric potential at a point outside the disk is given by the equation:
  • V = k * σ * (√(R^2 + x^2) - √(R^2 + a^2)) (where σ is the surface charge density, R is the radius of the disk, x is the distance from the center of the disk, and a is the radius of the Gaussian surface)

Example 8

• Problem:

  • A uniformly charged disk with a radius of 0.2 m has a surface charge density of 2 μC/m^2. Find the electric potential at a distance of 0.5 m from the center of the disk.

• Solution:

  • Given:
    • Radius of the disk, R = 0.2 m
    • Surface charge density, σ = 2 μC/m^2
    • Distance from center, x = 0.5 m
  • Calculating the potential:
    • V = k * σ * (√(R^2 + x^2) - √(R^2 + a^2)) (√ is the square root)

Summary

• Potential energy is calculated using: U = qV
• Electric potential difference is given by: ∆V = Vf - Vi
• Potential energy due to a system of charges: U = Σ (qi * V)
• Work done along an equipotential surface is zero
• Potential on an equipotential surface is given by: V = k * (q1 / r1 + q2 / r2 + q3 / r3 + …)
• Electric potential due to continuous charge distribution: V = k ∫(dq / r)

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Electric Potential due to a Line of Charge

• A line of charge is another example of continuous charge distribution.
• The electric potential at a point due to a line of charge can be calculated using the equation:
  • V = k * λ * ln(r2 / r1) (where λ is the linear charge density, r1 is the initial distance, and r2 is the final distance)

Example:

• Problem: A line charge with a linear charge density of 10 μC/m is stretched from a distance of 2 m to 5 m. Calculate the potential difference.
• Solution: Using the formula V = k * λ * ln(r2 / r1), we can calculate the potential difference.

Electric Potential due to a Charged Ring

• A charged ring is a 2-dimensional charge distribution.
• The electric potential at a point on the axis of a charged ring can be calculated using the equation:
  • V = k * Q / (√(R^2 + h^2)) (where Q is the total charge of the ring, R is the radius of the ring, and h is the distance from the center to the point on the axis)

Example:

• Problem: A charged ring has a total charge of 6 μC and a radius of 0.5 m. Calculate the electric potential at a point on the axis at a distance of 0.3 m from the center.
• Solution: Using the formula V = k * Q / (√(R^2 + h^2)), we can calculate the electric potential at the given point.

Electric Potential due to a Charged Plane

• A charged plane is another example of a 2-dimensional charge distribution.
• The electric potential at a point near a charged plane can be calculated using the equation:
  • V = E * d (where E is the electric field strength near the plane and d is the perpendicular distance from the plane)

Example:

• Problem: A charged plane creates an electric field strength of 1000 N/C near its surface. Calculate the electric potential at a distance of 2 cm from the plane.
• Solution: Using the formula V = E * d, we can calculate the electric potential at the given distance from the plane.

Capacitors and Electric Potential

• A capacitor is a device used to store electrical energy.
• The electric potential difference across a capacitor is denoted by ∆V and is given by the equation:
  • ∆V = Q / C (where Q is the charge stored in the capacitor and C is the capacitance of the capacitor)

Example:

• Problem: A capacitor stores a charge of 5 μC and has a capacitance of 3 μF. Calculate the electric potential difference across the capacitor.
• Solution: Using the formula ∆V = Q / C, we can calculate the electric potential difference.

Combining Capacitors in Series

• When capacitors are connected in series, the total capacitance is given by the equation:
  • 1 / C_total = 1 / C1 + 1 / C2 + 1 / C3 + … (where C1, C2, C3, … are the individual capacitances)

Example:

• Problem: Three capacitors with capacitances of 4 μF, 6 μF, and 8 μF are connected in series. Calculate the total capacitance.
• Solution: Using the formula 1 / C_total = 1 / C1 + 1 / C2 + 1 / C3, we can calculate the total capacitance.

Combining Capacitors in Parallel

• When capacitors are connected in parallel, the total capacitance is given by the equation:
  • C_total = C1 + C2 + C3 + … (where C1, C2, C3, … are the individual capacitances)

Example:

• Problem: Three capacitors with capacitances of 4 μF, 6 μF, and 8 μF are connected in parallel. Calculate the total capacitance.
• Solution: Using the formula C_total = C1 + C2 + C3, we can calculate the total capacitance.

Energy Stored in a Capacitor

• The energy stored in a capacitor is given by the equation:
  • U = 0.5 * C * (∆V)^2 (where U is the energy stored, C is the capacitance, and ∆V is the potential difference across the capacitor)

Example:

• Problem: A capacitor with a capacitance of 10 μF has a potential difference of 12V across its terminals. Calculate the energy stored in the capacitor.
• Solution: Using the formula U = 0.5 * C * (∆V)^2, we can calculate the energy stored.

Dielectrics and Capacitance

• A dielectric is an insulating material placed between the plates of a capacitor, which increases the capacitance.
• The capacitance of a capacitor with a dielectric material is given by the equation:
  • C = k * C0 (where C is the capacitance with the dielectric, C0 is the capacitance without the dielectric, and k is the dielectric constant)

Example:

• Problem: A capacitor without a dielectric has a capacitance of 8 μF. If a dielectric with a dielectric constant of 5 is inserted between the plates, calculate the new capacitance.
• Solution: Using the formula C = k * C0, we can calculate the new capacitance.

Stored Energy with a Dielectric

• The energy stored in a capacitor with a dielectric material is given by the equation:
  • U = 0.5 * C * (∆V)^2 (where U is the energy stored, C is the capacitance with the dielectric, and ∆V is the potential difference across the capacitor)

Example:

• Problem: A capacitor with a capacitance of 10 μF and a potential difference of 20V has a dielectric material with a dielectric constant of 4. Calculate the energy stored in the capacitor.
• Solution: Using the formula U = 0.5 * C * (∆V)^2, we can calculate the energy stored.

Conclusion

• In this lecture, we covered:
  • Electric potential due to continuous charge distributions
  • Potential on equipotential surfaces
  • Calculations of potential for line charges, rings, planes, and capacitors
  • Combinations of capacitors in series and parallel
  • Energy stored in capacitors with and without