Question

2. Three charges + 3q + q and Q are placed on a st. line with equal separation. In
order to maket the net force on q to be zero, the value of Q will be : (a) +3q (b)+2q(c) -3q
(d) -4q​

Answer 1

Answer:

a) +3q :O:O:O:O:O:O:O:O:O:O:O:O

Answer 2

Given: Three charges + 3q, + q and Q on straight line

Distance between all three charges is equal

To Find: The value of Q to make the net force on q to be zero

Solution:

Force acting between two charges placed on a straight line is given as:

$\[F = K\frac{{{q_1}{q_2}}}{{{r^2}}}\]$

Suppose, the distance between all the three charges is $r$.

Therefore, the force acting between charges + 3q and + q is

$\[{F_1} = K\frac{{3q \times q}}{{{r^2}}}\]$                    ... (i)

Similarly, the force acting between charges Q and + q is

$\[{F_2} = K\frac{{{\rm Q} \times q}}{{{r^2}}}\]$                    ... (ii)

In order to make the net force on q to be zero,

$\[{F_2} = {F_1}\]$

From equations (i) and (ii), we have,

$\[K\frac{{{\rm Q} \times q}}{{{r^2}}} = K\frac{{3q \times q}}{{{r^2}}}\]$

$\[Q = 3q\]$

Hence, in  order to make the net force on q to be zero, the value of Q will be  +3q.