Question

Two fixed point charges + 4 q and + 2 cube are separated by a distance x where should be the third point charge be placed for it to be in equilibrium?

Answer 1

Let it be placed at a distance y from 4Q and x-y from 2Q.
or equilibrium we have:
k4Qq/(y 2 )=k2Qq/(x-y) 2
Hence, 2(x-y) 2 =y 2
y 2 -4xy+2x 2 =0
Hence ,y= x(4-2(2) 0.5 )/2

Answer 2

Answer:

The third point charge should be placed at a distance - x ( 1 + - √ 2 ) for it to be in equilibrium.

Explanation:

The two fixed point charges are placed in an equilibrium, then a third charge is introduced, its position to let the system be in equilibrium can be calculated as follows -  

Given, the charge q1 = +4 q

Charge q2 = + 2 q

Charge q3 = q

Distance between q1 and q2 = x  

Distance between q1 and q3 = x - a

Distance between q2 and q3 = a

Now, to have equilibrium of the system, force on charge q1 and q2 exerted by q3 must be equal, so, we have –

Force on q3 by q1 = force on q3 by q2

$\begin{array}{l}{\frac{k \times 4 q \times q}{(x-a)^{2}}=\frac{k \times 2 q \times q}{a^{2}}} \\ {\frac{k \times 4 q \times q}{k \times 2 \times q}=\frac{(x-a)^{2}}{a^{2}}} \\ {2=\frac{(x-a)^{2}}{a^{2}}} \\ {2 a^{2}=(x-a)^{2}} \\ {2 a^{2}=x^{2}+a^{2}-2 a x} \\ {a^{2}+2 a x-x^{2}=0}\end{array}$

On solving this equation, we get –

a = - x ( 1 + - √ 2 )

So the charge q3 be placed at the distance of - x ( 1 + - √ 2 ) from the charge q2.