2019 A/L Physics Past Paper · Question 27 (English Medium)

Answer: (4) $\frac{Q}{4πε_0\sqrt{r^2 + d^2}}$

What the question asks

We want:

The electric potential at $P$ on the axis of a ring.

The ring has radius $r$ and a total charge $Q$ spread evenly. $P$ is at the distance $d$ from the centre.

Relevant theory

The potential at a distance $s$ from a point charge $q$ is:

$$V = \dfrac{q}{4\pi\varepsilon_0 s}$$

Potential is a scalar.

For many charges, add the potentials as plain numbers.

The electric potential at a point in an electric field is the work done taking a unit positive charge (+1 C) from infinity up to that point.

Required equations

$V_i = \frac{q_i}{4\pi\varepsilon_0 s_i}$

$V = \sum_i V_i$

Reasoning and calculation

The symbols:

$r$ is the radius of the ring.

$d$ is the distance of $P$ from the centre, along the axis.

$q_i$ is one small charge on the ring. All of them add to $Q$.

$s_i$ is the distance of $P$ from that charge.

Every point of the ring is at the same distance from $P$.

From the right-angled triangle:

$$s_i = \sqrt{r^2 + d^2}$$

It does not depend on $i$.

$V = \frac{1}{4\pi\varepsilon_0\sqrt{r^2 + d^2}}\sum_i q_i$

$V = \frac{Q}{4\pi\varepsilon_0\sqrt{r^2 + d^2}}$

Reading the options

Compare the form.

$\sqrt{r^2 + d^2}$ is in the denominator and there is no other factor.

That is option (4).

Checks

Check the dimensions.

$\dfrac{Q}{\varepsilon_0 \times \text{length}}$ has the form of the potential of a point charge.

Option (3) has length squared in the denominator. Option (5) fails the limit check below.

Check the limits.

At $d = 0$ (the centre) $V = \dfrac{Q}{4\pi\varepsilon_0 r}$. All the charge is at the distance $r$. That is right.

Far away, $d \gg r$, $V \to \dfrac{Q}{4\pi\varepsilon_0 d}$, like a point charge. That is right.

Option (5) becomes infinite as $d \to 0$. That is wrong.

Conclusion

$V = \frac{Q}{4\pi\varepsilon_0\sqrt{r^2 + d^2}}$

Every charge is at the distance $\sqrt{r^2 + d^2}$ from $P$.

So $V = \dfrac{Q}{4\pi\varepsilon_0\sqrt{r^2 + d^2}}$.

The answer is option (4).

Adding a factor $\cos\theta$, as when working out the field.

Potential is a scalar. It has no direction.

Components are needed only for the field, which has a direction.