2021 A/L Physics Past Paper · Question 45 (English Medium)

Answer: (2) Graph (2)

What the question asks

We want: the graph that shows how the electric field intensity $E$ changes with the distance $r$ from the centre.

The figure shows a metal sphere of radius $2R$ with a hollow of radius $R$. The sphere has a net charge $+q$. A point charge $+q$ is at the centre.

The plan: find $E$ in three regions, $r < R$, $R < r < 2R$ and $r > 2R$. Then match the graphs.

Relevant theory

It has been shown by scientist Michael Faraday that the magnitude of the two induced

charges at B and C here are separately equal to the magnitude of the inducing charge in A.

When a conducting sphere is given an electric charge, we already know that this will reside only on the outer surface of the sphere.

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.

Applying it to this question

Read the figure. The point charge $+q$ is at the centre of the cavity. The metal is the shaded ring between $R$ and $2R$.

Charges in the metal can move. They settle where the field in the metal is zero.

Variables we can read: the three regions.

Region 1, the cavity: $r < R$. Region 2, the metal: $R < r < 2R$. Region 3, outside: $r > 2R$.

Reasoning and calculation

Region 1, the cavity ($r < R$). Only the point charge $+q$ makes the field.

$$E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$$

It falls as $r$ grows. At $r = R$ it is $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{R^2}$. This is not zero.

Region 2, the metal ($R < r < 2R$). The field is zero. For that, the cavity wall takes an induced charge $-q$, equal to the charge $+q$ that induces it.

The sphere has a net charge $+q$. So the outer surface holds $q + q = 2q$.

Region 3, outside ($r > 2R$). The sphere acts as if its charge sits at the centre. With the point charge, the total is $2q$.

$$E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2q}{r^2}$$

At $r = 2R$ this is $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{2R^2}$. That is half of the field just inside $R$.

Reading the options

The graph must fall to a value at $R$, stay at zero up to $2R$, then jump back to half that value and fall again.

Only graph (2) does this. It is option (2).

Why the other options fail

The field outside, at $2R$, is half of the field at $R$. In graph (1) the two are equal.

The field in the cavity comes from the point charge. It does not fall to zero at $R$.

The field outside, at $2R$, comes from the charge $2q$, not $q$. It is half of the field at $R$, not a quarter.

The charge in the cavity is a point at the centre. Its field is not zero at the centre and does not grow in a straight line.

Checks

Does it make sense?

Far away, the field should look like that of one charge $2q$ at the centre. Graph (2) falls as $\dfrac{1}{r^2}$ outside, with the strength of $2q$.

The jump at $2R$ is the surface charge $2q$, and the dip to zero in the metal is the shielding.

$(2 * q / ((2 * r) * (2 * r))) / (q / (r * r))$

Conclusion

$E$ falls as $\dfrac{1}{r^2}$ up to $R$, is zero from $R$ to $2R$, then jumps to half of the value at $R$ and falls again.

This is graph (2). The answer is (2).

In a conductor with a cavity, find the charge inside, the induced charge on the cavity wall and the charge on the outer surface.

Outside, only the total charge counts: $2q$ here.

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