Capacitance of an Isolated Sphere (OCR A Level Physics)

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Capacitance of an Isolated Sphere

  • The capacitance, C, of a charged sphere, is defined as the charge per unit potential at the surface of the sphere

C equals Q over V

  • Where:
    • C = capacitance (F)
    • Q = charge (C)
    • V = potential difference (V)

  • The charge on the surface of a spherical conductor can be considered as a point charge at its centre
  • The potential V of an isolated point charge is given by:

V equals fraction numerator Q over denominator 4 straight pi epsilon subscript italic 0 R end fraction

  • Where:
    • R = radius of sphere (m)
    • ε0 = permittivity of free space
  • The charge, Q, is not the charge of the capacitor itself, it is the charge stored on the surface of the spherical conductor
  • Combining these equations gives an expression for the capacitance of an isolated sphere:

V equals fraction numerator Q over denominator 4 straight pi epsilon subscript italic 0 R end fraction equals Q over C

C = 4πε0R

Worked example

Lightning can be simulated in a laboratory using an isolated metal sphere to investigate electrical discharge.

A sphere of radius 75 cm is charged to a potential of 1.5 MV.

Following the electrical discharge, the sphere loses 95% of its energy.

Calculate:

a)
The capacitance of the sphere.
b)
The potential of the sphere after discharging.

Part (a)

Step 1: List the known quantities

    • Radius of sphere, R = 75 cm = 75 × 10−2 m
    • Permittivity of free space, ε0 = 8.85 × 10−12 F m−1

Step 2: Write out the equation for the capacitance of a charged sphere

C = 4πε0R

Step 3: Calculate the capacitance

C = 4π × (8.85 × 10−12) × (75 × 10−2)

C = 8.34 × 10−11 F

Part (b)

Step 1: List the known quantities

    • Original potential, V1 = 1.5 MV = 1.5 × 106 V
    • Final potential = V2
    • Original energy = E1
    • Final energy, E2 = 0.05 E1

Step 2: Write out the equation for the energy stored by a capacitor

E equals 1 half C V squared

Step 3: Write out equations for energy before and after discharge

E subscript 1 equals 1 half C V subscript 1 squared

E subscript 2 equals 1 half C V subscript 2 squared

Step 4: Equate the two expressions and simplify

    • Since E2 = 0.05 E1

1 half C V subscript 2 squared equals 0.05 open parentheses 1 half C V subscript 1 squared close parentheses

V subscript 2 squared space equals space 0.05 open parentheses V subscript 1 squared close parentheses

V subscript 2 space equals space square root of 0.05 end root space V subscript 1

Step 5: Calculate the final potential, V2

V2square root of 0.05 end root × (1.5 × 106) = 3.35 × 105 V

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Katie M

Author: Katie M

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

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