2018 A/L Physics Past Paper · Question 45 (English Medium)

Answer: (1) 1 cm

What the question asks

A charge is brought from infinity to $O$ with no work.

For that, the potential at $O$ must be zero.

We want: the distance $r$ of $-q$ that makes it zero.

Relevant theory

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.

The work to bring a charge $q'$ from infinity to $O$ is

$$W = q'V_O$$

If the work is zero, then $V_O = 0$.

Potential has no direction. So we add the share of each charge with its sign.

$$V = \frac{kQ}{r}$$

Applying it to this question

The charges that add to the potential at $O$:

$+q$ at 2 cm, $+q$ at 3 cm, $+q$ at 6 cm.

$-q$ at $r$.

$V_1 = \frac{kq}{2}$

$V_2 = \frac{kq}{3}$

$V_3 = \frac{kq}{6}$

$V_4 = -\frac{kq}{r}$

$+q$, at 2 cm

A positive charge. Its share at $O$ is positive.

It is the nearest. So its share is the largest.

$+q$, at 3 cm

A positive charge. Its share is positive.

It is smaller than the share of the charge at 2 cm.

$+q$, at 6 cm

A positive charge. It is the farthest away.

So its positive share is the smallest.

$-q$, at $r$

A negative charge. Its share is negative.

It must cancel the other three exactly.

Reasoning and calculation

The positive charges give positive shares. The negative charge gives a negative share.

Potential has no direction.

So it does not matter where the charges are. Only the distance matters.

$V_O = kq\left(\frac12 + \frac13 + \frac16 - \frac1r\right)$

$1$

$\frac12 + \frac13 + \frac16 = \frac{3 + 2 + 1}{6}$

$1 - \frac{1}{r} = 0$

$r = 1$

Reading the options

The options are 1 cm to 5 cm.

Only $r = 1$ cm gives $V_O = 0$.

Why the other options fail

At $r = 2$ cm, $V_O = \frac{kq}{2}$. This is not zero.

At $r = 3$ cm, $V_O = \frac{2kq}{3}$. This is not zero.

At $r = 4$ cm, $V_O = \frac{3kq}{4}$. This is not zero.

At $r = 5$ cm, $V_O = \frac{4kq}{5}$. This is not zero.

Checks

$1 / (1/2 + 1/3 + 1/6)$

The directions of the charges are not given.

They are not needed, because potential has no direction.

If the force or the field were asked, the directions would matter.

Conclusion

$r = 1\ \text{cm}$

$$\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1 = \frac{1}{r}$$

So $r = 1$ cm.

The answer is option (1).

Treating potential like a force and looking for directions.

Potential has no direction. Add $\frac{kq}{r}$ for each charge, with its sign.