10.2.4 Potential in a Charged Sphere

Potential in a Charged Sphere

The electric potential in the field due to a point charge is defined as:

V = \frac{Q}{4 \times π \times ε_{0} \times r}

- Where:
  - V = the electric potential (V)
  - Q = the point charge producing the potential (C)
  - ε₀ = permittivity of free space (F m^-1)
  - r = distance from the centre of the point charge (m)

This equation shows that for a positive (+) charge:
- As the distance from the charge r decreases, the potential V increases
- This is because more work has to be done on a positive test charge to overcome the repulsive force
For a negative (−) charge:
- As the distance from the charge r decreases, the potential V decreases
- This is because less work has to be done on a positive test charge thanks to the effect of the attractive force

Unlike the gravitational potential equation, the minus sign in the electric potential equation will be included in the charge
The electric potential changes according to an inverse square law with distance

Potential around charged sphere, downloadable AS & A Level Physics revision notes

The potential changes as an inverse law with distance near a charged sphere

Note: this equation still applies to a conducting sphere. The charge on the sphere is treated as if it concentrated at a point in the sphere from the point charge approximation

Worked Example

A Van de Graaf generator has a spherical dome of radius 15 cm. It is charged up to a potential of 240 kV. Calculate:

a) The charge stored on the dome

b) The potential at a distance of 30 cm from the dome

Part (a)

Step 1: Write down the known quantities

- - Radius of the dome, r = 15 cm = 15 × 10^-2 m
  - Potential difference, V = 240 kV = 240 × 10³ V

Step 2: Write down the equation for the electric potential due to a point charge

V = \frac{Q}{4 \times π \times ε_{0} \times r}

Step 3: Rearrange for charge Q

$Q = V \times 4 \times π \times ε_{0} \times r$

Step 4: Substitute in values

$Q = (240 \times 10^{3}) \times (4 \times π \times 8.85 \times 10^{- 12}) \times (15 \times 10^{- 2}) = 4.0 \times 10^{- 6} C = 4.0 μ C$

Part (b)

Step 1: Write down the known quantities

- Q = charge stored in the dome = 4.0 μC = 4.0 × 10^-6 C
- r = radius of the dome + distance from the dome = 15 + 30 = 45 cm = 45 × 10^-2 m

Step 2: Write down the equation for electric potential due to a point charge

V = \frac{Q}{4 \times π \times ε_{0} \times r}

Step 3: Substitute in values

V = \frac{(4.0 \times 10^{- 6})}{(4 \times π \times 8.85 \times 10^{- 12}) \times (45 \times 10^{- 2})} = 79.93 \times 10^{3} = 80 k V

Worked Example

A metal sphere of diameter 15 cm is negatively charged. The electric field strength at the surface of the sphere is 1.5 × 10⁵ V m^-1. Determine the total surface charge of the sphere.

Step 1: Write down the known values

- Electric field strength, E = 1.5 × 10⁵ V m^-1
- Radius of sphere, r = 15 / 2 = 7.5 cm = 7.5 × 10^-2 m

Step 2: Write out the equation for electric field strength

V = \frac{Q}{4 \times π \times ε_{0} \times r}

Step 3: Rearrange for charge Q

Q = V \times 4 \times π \times ε_{0} \times r

Step 4: Substitute in values

Q = (4 \times π \times 8.85 \times 10^{- 12}) \times (1.5 \times 10^{5}) \times {(7.5 \times 10^{- 2})}^{2} = 9.38 \times 10^{- 8} C = 94 n C

DP IB Physics: HL

Revision Notes

10.2.4 Potential in a Charged Sphere

Potential in a Charged Sphere

Worked Example

Worked Example

Author: Katie

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