An isolated solid conducting sphere is initially uncharged - AQA - A-Level Physics - Question 3 - 2022 - Paper 2

To find the electric field strength E at a distance of 0.30 m, we first need to determine the change in electric potential ΔV over the relevant distance Δr.

From Figure 3, at r = 0.30 m, the electric potential V is approximately -1.2 × 10^6 V.

Thus, taking two points around this distance,

• V1 at r = 0.20 m ≈ -0.8 × 10^6 V
• V2 at r = 0.30 m ≈ -1.2 × 10^6 V

The change in potential is:

$$ΔV = V2 - V1 = (-1.2 imes 10^6) - (-0.8 imes 10^6) = -0.4 imes 10^6 ext{ V}$$

The change in distance Δr = 0.30 m - 0.20 m = 0.10 m.

Now we can find the electric field strength:

E = -rac{ΔV}{Δr} = -rac{-0.4 imes 10^6}{0.10} = 4.0 imes 10^6 ext{ N/C}.

The SI unit for electric field strength is N/C.

The capacitance C of a conductor can be calculated using the formula:

C = rac{Q}{V}

Where Q is the charge and V is the potential. From earlier calculations, we can assume the charge Q at the surface of the sphere can be estimated. Considering a spherical conductor, the capacitance is also related to its radius by:

$$C = 4 ext{π} imes ext{ε}_0 imes r$$

Given the radius r of the sphere is 0.10 m, and where ε₀ (the permittivity of free space) is approximately 8.85 × 10^-12 F/m, we have:

$$C = 4 imes ext{π} imes (8.85 imes 10^{-12}) imes 0.10 ext{ m} \ ≈ 1 imes 10^{-11} ext{ F}.$$

The energy stored (U) in a capacitor is given by:

U = rac{1}{2} C V^2

Using the capacitance found earlier (C = 1 × 10^{-11} F) and the new potential V = 1.0 × 10^6 V:

U = rac{1}{2} imes (1 imes 10^{-11}) imes (1.0 imes 10^6)^2

Calculating this gives:

U = rac{1}{2} imes (1 imes 10^{-11}) imes (1 imes 10^{12}) \ = 0.5 ext{ J}.

To find the change in energy, we need to compare this with the energy at the previous potential if it was calculated. Assuming the previous energy was higher, the change in energy is:

$$ext{Change in energy} = ext{Final energy} - ext{Initial energy} = 0.5 - ( ext{initial calculation if provided}).$$