Question

The distance between two fixed point charges 4e and e is how shuld a third charge

Answer 1

Given,

qA = +4e

qB = +e

AB = x

Let, Q be the third point charge located at some point P.

AP = x

therefore, PB = (a-x)

Force on Q due to qA,

F1 = kqAQ/x2

Force on Q due to qB

F2= kQqB/(a-x)2

Since the charge is in equilibrium,

F1 = F2

kqAQ/x2 = k qBQ/(a-x)2

4e/x2 = e/(a-x)2

4/x2 = 1/(a-x)2 (Taking square root on both sides)

2/x = 1/(a-x)

2(a-x) = x(1)

2a - 2x = x

2a = 3x

x = 2/3a

Therefore, the third point charge Q should be placed 2/3a from A.