2021 A/L Physics Past Paper · Question 17 (English Medium)

Answer: (1) $x = y$

Given information

We know:

The charges $+4q$ and $-q$ are fixed, a distance $x$ apart.

A third charge $+q$ sits on the same line, a distance $y$ from $-q$, on the side away from $+4q$.

The net force on $+q$ is zero.

What the question asks

We want: the relation between $x$ and $y$.

The tie: the net force on $+q$ is zero. So the force from $+4q$ and the force from $-q$ must be equal, and opposite.

Relevant theory

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

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

Required equations

$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_1Q_2}{r^2}$

$F_{4q} = F_{-q}$

Applying it to this question

Apply the law to each charge.

The charge $+4q$ repels $+q$, because they are like charges. It pushes $+q$ away. The distance between them is $x + y$.

$$F_{4q} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{4q \times q}{(x + y)^2}$$

The charge $-q$ attracts $+q$, because they are unlike charges. It pulls $+q$ back. The distance is $y$.

$$F_{-q} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q \times q}{y^2}$$

$\dfrac{4}{(x + y)^2} = \dfrac{1}{y^2}$

$4y^2 = (x + y)^2$

$2y = x + y$

$x = y$

Reasoning and calculation

Directions first. The push and the pull point opposite ways, so they can cancel.

If $+q$ were between the two charges, both forces would point the same way. They could never cancel.

Put $F_{4q} = F_{-q}$ and cancel the common factor:

$$\dfrac{4}{(x + y)^2} = \dfrac{1}{y^2}$$

Cross-multiply:

$$4y^2 = (x + y)^2$$

Take the square root of both sides. Both $x + y$ and $y$ are positive:

$$2y = x + y$$

So $x = y$. This is option (1).

Checks

Does it make sense?

Put $x = y = 1$. Then $+q$ is $2$ from $+4q$ and $1$ from $-q$.

The push is $\dfrac{4}{2^2} = 1$ and the pull is $\dfrac{1}{1^2} = 1$. They are equal, so the forces cancel.

$4 * y * y / ((x + y) * (x + y))$

Conclusion

$x = y$

The forces on $+q$ cancel when $4y^2 = (x + y)^2$, which gives $x = y$.

The answer is (1).

For a point with no net force, write the two forces and set them equal.

Measure each distance from the charge you are working on. Here $+4q$ is $x + y$ away, not $x$.

Exam tip