2019 A/L Physics Past Paper · Question 29 (English Medium)

Answer: (2) Graph (2)

What the question asks

We want:

How the electric field $E$ changes with the distance $x$ from the centre.

The inner shell is at the potential $V$. The outer shell is earthed.

Then choose the matching graph.

Relevant theory

Gauss's law: the flux through a closed surface is the enclosed charge divided by $\varepsilon_0$.

For a sphere of radius $x$ round the centre:

$$E \cdot 4\pi x^2 = \dfrac{Q_{enc}}{\varepsilon_0}$$

The inner shell carries $+Q$.

The earthed outer shell gets $-Q$ on its inside face and nothing on its outside face.

Since there is no charge inside the sphere the field intensity inside too is zero.

Required equations

$E \cdot 4\pi x^2 = \frac{Q_{enc}}{\varepsilon_0}$

$E = \frac{Q}{4\pi\varepsilon_0 x^2}$

Reasoning and calculation

The axes:

Across is the distance $x$ from the centre.

Up is the size of the field $E$, which is zero or more.

Start at the centre.

From $x = 0$ to $x < r$ no charge is enclosed.

So $E = 0$.

Three regions:

(i) $x < r$, inside the inner shell.

(ii) $r < x < R$, between the shells.

(iii) $x > R$, outside.

The behaviour changes at $x = r$ and at $x = R$.

Each region:

(i) $Q_{enc} = 0$, so $E = 0$.

(ii) $Q_{enc} = +Q$, so $E = \dfrac{Q}{4\pi\varepsilon_0 x^2}$. It falls as $x$ grows.

(iii) $Q_{enc} = Q - Q = 0$, so $E = 0$.

The values at the edges.

At $x = r$ the field jumps from 0 to its highest value, $\dfrac{Q}{4\pi\varepsilon_0 r^2}$.

At $x = R$ just inside, $E = \dfrac{Q}{4\pi\varepsilon_0 R^2}$, which is above zero.

Just outside $R$ it drops to 0 at once.

The shape.

From $r$ to $R$ it is a curve falling as $\dfrac{1}{x^2}$.

It is on zero on both sides.

At $R$ the curve drops suddenly. It does not first reach zero.

Reading the options

Graph (2) matches.

It falls from its peak at $r$ and drops suddenly from a value above zero at $R$.

It is zero on both sides.

Why the other options fail

The curve reaches zero at $R$.

The field is constant.

The field grows inside the inner shell.

A constant field inside the inner shell.

Conclusion

$E$ is zero for $x < r$ and for $x > R$.

Between the shells it falls as $\dfrac{1}{x^2}$ and drops to zero suddenly at $R$.

That is graph (2).

Thinking that the field falls gradually to zero as it nears the earthed shell.

It is the potential that falls gradually to zero at $R$, not the field.

The field changes suddenly where it crosses the layer of charge $-Q$.