2022 A/L Physics Past Paper · Question 29 (English Medium)

Answer: (5) $+2q$

Given information

What we know:

$X$ has the charge $+q$ and is a distance $x$ to the left of $P$. $Y$ is a distance $\sqrt{2}\,x$ to the right of $P$.

A small negative charge $-q_0$ placed at $P$ stays at rest. So the two forces on it cancel.

The question has no numbers. So we take the size of $q$ and $x$ as $1$ unit each and work in those units.

What the question asks

We want: the charge $Q$ of $Y$, with its sign.

What each letter means:

$q$ ($\text{C}$): the charge of $X$.

$Q$ ($\text{C}$): the charge of $Y$.

$q_0$ ($\text{C}$): the size of the negative charge at $P$.

$x$ ($\text{m}$): the distance from $X$ to $P$.

$F_X$, $F_Y$ ($\text{N}$): the forces on the charge at $P$ from $X$ and from $Y$.

$k$ ($\text{N m}^2\,\text{C}^{-2}$): the constant in Coulomb's law.

Relevant theory

Like charges (+, + and -, -) repel each other while unlike charges (+, -) attract each other.

The mutual force between two point electric charges is directly proportional to the product of the two charges and inversely proportionatel to the square of the distance between them

For example, two lamps light a table. A weak lamp close to the table and a strong lamp far from it can give the same light. A farther source must be stronger to have the same effect.

The same happens here. The force from a charge falls as the distance grows. So the farther charge must be bigger to pull as hard.

$$F = k\,\dfrac{Q_1 Q_2}{r^2}$$

In words: the force is in proportion to the product of the charges, and in inverse proportion to the square of the distance.

Direction first. The charge at $P$ is negative, and $X$ is positive, so $X$ attracts it to the left. For the charge to stay at rest, $Y$ must pull it to the right. A pull towards $Y$ means $Y$ attracts it, so $Y$ is positive.

Size next. The two forces are equal.

$$k\,\dfrac{q\,q_0}{x^2} = k\,\dfrac{Q\,q_0}{(\sqrt{2}\,x)^2}$$

In words: the pull of $X$ equals the pull of $Y$.

Required equations

$F_X = k\,\dfrac{q\,q_0}{x^2}$

$F_Y = k\,\dfrac{Q\,q_0}{(\sqrt{2}\,x)^2}$

$F_X = F_Y \Rightarrow Q = q\,\dfrac{(\sqrt{2}\,x)^2}{x^2}$

Applying it to this question

The charges and distances appear as ratios, so $k$ and $q_0$ cancel. The answer is a number times $q$, and it has the unit of charge.

$1$

$F_X = \dfrac{q}{x^2} = \dfrac{1}{1} = 1$

$0.5$

$F_Y = \dfrac{Q}{(\sqrt{2}\,x)^2} = \dfrac{Q}{2}$

$2$

$\dfrac{Q}{2} = 1 \Rightarrow Q = 2$

Reasoning and calculation

$Q = +2q$. The sign is positive because $Y$ has to attract the negative charge. The size is $2q$, since $Y$ is farther away and $(\sqrt{2})^2 = 2$.

That is option (5).

Checks

$q * y * y / (x * x)$

Does it make sense?

$Y$ is farther from $P$ than $X$ is. A farther charge needs to be bigger to pull as hard. So $Q$ must be bigger than $q$, and $2q$ is.

It must also attract the negative charge, so it is positive. Only the options $+\dfrac{q}{2}$, $+\dfrac{q}{\sqrt{2}}$ and $+2q$ are positive, and only $+2q$ is bigger than $q$.

Conclusion

$Q = +2q$

$Y$ must attract the negative charge, so it is positive. The equal pulls give $Q = q \times \dfrac{(\sqrt{2}\,x)^2}{x^2} = 2q$.

The answer is (5).

A charge at rest means the forces on it cancel. Find the directions first, then make the sizes equal.

The farther charge must be bigger, by the square of the ratio of the distances.

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