2024 A/L Physics Past Paper · Question 22 (English Medium)

Answer: (2) Graph (2)

What the question asks

This question is about the electric potential around a hollow metal ball that carries charge.

The charge sits only on the outer surface of the metal. We walk from the centre of the ball, out through its surface, and far away.

What we must find: which graph shows how the potential $V$ changes with the distance $r$ from the centre.

How we find it: split the walk into two parts, inside the shell and outside it. Find what $V$ does in each part, then join them at the surface.

What each letter means:

$Q$ ($\text{C}$): charge on the shell.

$R$ ($\text{m}$): radius of the shell.

$r$ ($\text{m}$): distance from the centre.

$V$ ($\text{V}$): electric potential at that distance.

$E$ ($\text{N C}^{-1}$): electric field intensity.

Relevant theory

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.

Thus starting from the surface of the sphere, the potential towards the inside remains unchanged and is constant.

What is happening: outside the shell, the charge acts as if it were all at the centre. So the potential falls off as $1/r$ as we move away.

Inside the shell there is no charge, so there is no field. With no field, nothing changes the potential as we move about inside. It stays at the value it has on the surface.

For example, a car caught in a thunderstorm. If lightning hits it, the charge runs over the outside of the metal body.

Inside, everything stays at the same potential, so no current flows through the people sitting there.

In words: inside a charged metal shell the potential is the same everywhere.

Required equations

$V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}$

$V_0=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}$

$V=V_0$

Reasoning and calculation

Two regions: inside the shell, from $r=0$ to $r=R$, and outside it, from $r=R$ outwards.

Inside, $V$ is flat at $V_0$. Outside, $V$ falls as $1/r$. The two parts meet at $r=R$, where both give $V_0$.

$V_0$

$V(0)$

$V(R)$

$\dfrac{V_0}{2}$

$V(2R)=\dfrac{R}{2R}V_0$

$0$

$V(\infty)$

The shape: a flat line at $V_0$ from the centre to the surface. From the surface outwards it curves down, falling fast at first and then more slowly, towards zero. There is no jump and no peak.

The graph we expect: flat from $r=0$ to $r=R$, then a falling curve that tends to zero.

$1/x$

Does it make sense? The potential gradient is the field. Inside, the field is zero, and the flat line has zero slope. Outside, the field points away from a positive shell, and the curve slopes downwards. Both fit.

Reading the options

Now match it with the five graphs.

Graph (2) is flat from the axis and then falls as a curve. It matches.

Graph (1) is zero inside, (3) rises inside, (4) rises outward and (5) shoots up at the centre. None of these match.

Why the other options fail

Graph (1) is zero inside. Inside, $V$ stays at $V_0$.

Graph (3) rises inside. There is no field inside to make it rise.

Graph (4) rises as $r$ grows. $V$ falls away from the shell.

Graph (5) is a point charge. Inside the shell $V$ stops at $V_0$.

Conclusion

$V_0=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}$

The potential is constant inside the shell and falls as $1/r$ outside. That is graph (2).

The answer is (2).

Inside a charged conductor, $E=0$, so $V$ is flat and equal to the surface value.

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