2021 A/L Physics Past Paper · Question 47 (English Medium)

Answer: (2) $\dfrac{Q^2}{8\pi\varepsilon_0 r}$

Given information

We know:

The big sphere has radius $2r$ and charge $+3Q$. The small sphere has radius $r$ and no charge.

They are touched, then moved apart. The surfaces are then $r$ apart. The system is in free space.

What the question asks

We want: the electrical potential energy of the two spheres after they are apart.

The plan: first share the charge between the spheres, then find the distance between the centres, then use the energy of a pair of charges.

Relevant theory

When the two spheres are connected by a conductor they reach the same potential V

It is known that in this case, with respect to point P, the charge Q on the sphere acts as if it were concentrated at the centre of the sphere.

The total potential energy of a system consisting of a number of charges would be the sum of the potential energies of all positive pairs of charges in the system.

Required equations

$V_1 = V_2$

$V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{R}$

$U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{d}$

Applying it to this question

$q_1 = 2q_2$

$Q$

$q_2 = \dfrac{3Q}{3}$

$2Q$

$q_1 = 3Q - Q$

$4r$

$d = 2r + r + r$

$\dfrac{Q^2}{8\pi\varepsilon_0 r}$

$U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2Q \times Q}{4r}$

$\dfrac{Q^2}{\varepsilon_0 r}$ has units $\dfrac{\text{C}^2}{\text{C}^2\,\text{N}^{-1}\text{m}^{-2}\ \text{m}} = \text{N m} = \text{J}$. So the answer is an energy.

Reasoning and calculation

The charge is shared as $2Q$ and $Q$. The centres are $4r$ apart.

The spheres are uniform, so each acts like a point charge at its centre. The energy of the pair is $\dfrac{Q^2}{8\pi\varepsilon_0 r}$. This is option (2).

Checks

$(a * b) / d$

Does it make sense?

Both spheres are positive, so they repel, and the energy is positive.

Charge was only shared, never made: $2Q + Q = 3Q$. The distance is between the centres, not between the surfaces.

Conclusion

$U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{d} = \dfrac{Q^2}{8\pi\varepsilon_0 r}$

The charges are $2Q$ and $Q$, and the centres are $4r$ apart.

$U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2Q \times Q}{4r} = \dfrac{Q^2}{8\pi\varepsilon_0 r}$. The answer is (2).

Touching conductors reach the same potential, so the charge is shared in the ratio of the radii.

For the energy, measure the distance from centre to centre.

Exam tip