Question

Two fixed point charges + 4Q and + 2 units are
separated by a distance x. Where should the third point
charge be placed for it to be in equilibrium ?

Answer 1

$\huge\underline{\overline{\mid{\bold{\pink{ANSWER-}}\mid}}}$

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/x^2

Force on Q due to qB

F2= kQqB/(a-x)^2

Since the charge is in equilibrium,

F1 = F2

kqAQ/x^2 = 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