Q3.15 Zero net field (K)

Two particles are fixed to a horizontal axis, one of charge +q at +2 cm from the origin, the other of charge -4q at +7 cm from the origin.

At what coordinate on the axis is the field produced by them equal to zero?

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Hint

The field will be zero where the electrostatic forces on a test positive charge are equal and opposite.
 

Solution

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Solution

Let’s identify the fixed charges as q1 has charge +q, x=2cm and q2 has charge -4q at x=7cm.

We imagine at test positive charge at x=L; the repulsive force between it and q1 is given by

$$F_- = - K \frac{q^2}{(L-2)^2}$$

and the attractive force between the test charge and q2 is

$$F_+ = K \frac{4q^2}{(L-7)^2}$$

Equating these gives:

$$ \frac{(L-7)^2}{(L-2)^2} = 4$$

You can just take square roots here, as in Q3.13

or if you prefer, expand and factorise:

L2-14L+49=4L2-16L+16

3L2-2L-33=0(3L-11)(L+3)=0

So L = -3cm or $\frac{11}{3} cm$

The value where the Electric field is zero is given by L=-3cm. (Think about why the other root cannot be correct)

(Note you can check your working for this using the Java applet in S2.12 )