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

To find the electric field strength (E), we first need to calculate the gradient of the electric potential (V) at r = 0.30 m. From the graph, we approximate:

At r = 0.30 m, V ≈ -1.2 × 10^6 V and at a nearby point, say r = 0.25 m, V ≈ -1.5 × 10^6 V.

Now, we calculate:

[ E = -\frac{\Delta V}{\Delta r} = -\frac{(-1.2 \times 10^6) - (-1.5 \times 10^6)}{0.30 - 0.25} = -\frac{0.3 \times 10^6}{0.05} = -6.0 \times 10^6 , \text{N/C} ]

Therefore, the electric field strength = 6.0 × 10^6 N/C.

The capacitance (C) of a sphere can be calculated using the formula:

[ C = \frac{Q}{V} ]

Where Q is the charge and V is the potential. For a sphere of radius 0.10 m:

Using the equation ( C = 4\pi\epsilon_0 r ), where ( \epsilon_0 \approx 8.85 \times 10^{-12} F/m ), we calculate:

[ C = 4\pi(8.85 \times 10^{-12})(0.10) = 1.11 \times 10^{-12} , F ]

Thus, it simplifies to approximately 1 × 10^-11 F.

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

[ U = \frac{1}{2} C V^2 ]

Substituting the capacitance calculated earlier and the given potential of 1.0 × 10^6 V:

[ U = \frac{1}{2} \times (1 \times 10^{-11}) \times (1.0 \times 10^6)^2 ] [ U = \frac{1}{2} \times (1 \times 10^{-11}) \times (1 \times 10^{12}) ] [ U = \frac{1}{2} \times (10.0) \approx 5.0 imes 10^0 J = 5.0 ext{ J} ]

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