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

At a distance of 0.30 m, we can find the electric field strength by determining the gradient from the graph in Figure 3. The change in potential ΔV can be observed from the graph between r = 0.2 m and r = 0.3 m:

• At r = 0.2 m, V ≈ -0.8 × 10⁶ V
• At r = 0.3 m, V ≈ -1.2 × 10⁶ V

Thus, ΔV = V_final - V_initial = (-1.2 × 10⁶) - (-0.8 × 10⁶) = -0.4 × 10⁶ V.

The change in distance Δr = 0.3 - 0.2 = 0.1 m.

Now, applying the formula:

$E = -\frac{\Delta V}{\Delta r} = -\frac{-0.4 \times 10^6}{0.1} = 4.0 \times 10^6 \text{ N/C}.$

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

For a sphere, the capacitance can be calculated using the equation:

$C = \frac{Q}{V}$

Where Q is the charge and V is the electric potential. To find the charge, we can use the relationship from part 0.3.2. Assuming Q is approximately 2.0 × 10⁻¹⁹ C:

Using the electric potential from the previous part, 1.0 × 10⁶ V:

$C = \frac{2.0 \times 10^{-19}}{1.0 \times 10^6} = 2.0 \times 10^{-25} F.$

However, the capacitance value for a spherical conductor is generally expressed as:

$C = 4\pi\epsilon_0 r$

Where $\epsilon_0 \approx 8.85 \times 10^{-12} F/m$ and r = 0.1 m, resulting in:

$C \approx 4\pi(8.85 \times 10^{-12})0.1 \approx 1 \times 10^{-11} F.$

The energy stored in a capacitor is given by the equation:

$E = \frac{1}{2} C V^2$

Using the value of capacitance C ≈ 1 × 10⁻¹¹ F and V = 1.0 × 10⁶ V:

$E = \frac{1}{2} (1 \times 10^{-11}) (1.0 \times 10^6)^2 = \frac{1}{2} (1 \times 10^{-11}) (1.0 \times 10^{12}) = 5.0 \text{ J}.$

Therefore, the change in energy stored by the sphere is 5.0 J.