2019 A/L Physics Past Paper · Question 17 (English Medium)

Answer: (1) $\dfrac{2q}{4\pi\varepsilon_0 a^2}$ →

What the question asks

We want:

The size and the direction of the electric field at $A$.

Eight point charges are around $A$ on a square grid.

Relevant theory

The field of a point charge $Q$ at a distance $r$ is:

$$E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$$

It points away from a positive charge and towards a negative charge.

With many charges, the fields add as vectors.

Also, field intensity is a vector quantity and its direction is represented by the direction of the force acting as this positive charge.

Applying it to this question

The sources, seen from $A$:

$+3q$ above, at $a$.

$+3q$ below, at $a$.

$+q$ on the left, at $a$.

$-q$ on the right, at $a$.

Four $+q$ at the corners, each at $\sqrt{2}\,a$.

Signs:

$+x$ to the right and $+y$ up.

A positive charge pushes the field away from it. A negative charge pulls it towards it.

Reasoning and calculation

Upper and lower $+3q$

$E_y = 0$

Each gives $\dfrac{3q}{4\pi\varepsilon_0 a^2}$.

The upper one points down. The lower one points up.

They are equal and opposite. They cancel.

The four corner $+q$

$\vec E_{corner} = 0$

Each gives $\dfrac{q}{4\pi\varepsilon_0 (\sqrt{2}a)^2} = \dfrac{q}{8\pi\varepsilon_0 a^2}$.

Opposite corners are at equal distances on opposite sides of $A$.

They cancel in pairs.

The left $+q$

$E = \frac{q}{4\pi\varepsilon_0 a^2}\ \rightarrow$

Its field points away from $+q$, to the right.

Its size is $\dfrac{q}{4\pi\varepsilon_0 a^2}$.

The right $-q$

Its field points towards $-q$, also to the right.

Add them as vectors.

The vertical parts cancel.

The corners give zero.

The two horizontal parts point the same way, to the right.

$E = \frac{q}{4\pi\varepsilon_0 a^2} + \frac{q}{4\pi\varepsilon_0 a^2} = \frac{2q}{4\pi\varepsilon_0 a^2}$

Reading the options

The size is $\dfrac{2q}{4\pi\varepsilon_0 a^2}$ and the direction is to the right.

That is option (1).

Option (3) has the same size but the opposite direction.

Why the other options fail

This option has a vertical field. There is none.

The direction is reversed.

This option does not cancel the $+3q$ charges.

Checks

A check by symmetry.

If $-q$ were $+q$, the picture would be fully symmetric. The field at $A$ would be zero.

Our sum would then give $q - q = 0$. So only the $-q$ breaks the symmetry.

Conclusion

$E = \frac{2q}{4\pi\varepsilon_0 a^2}\ \rightarrow$

The symmetric pairs cancel.

The left $+q$ and the right $-q$ both give a field to the right.

So the field is $\dfrac{2q}{4\pi\varepsilon_0 a^2}$ to the right.

The answer is option (1).

When the field of many charges is asked, first find the pairs that cancel.

Then work out only what is left.

Exam tip