Question

Two charges of same magnitude but of opposite nature are placed by some distance
away. F newton force is acting between them. If 50% of one charge is transfered
to another charge. Then force on them will be :-​

Answer 1

Explanation:

will remain same as the net cahrge is conserved...

Answer 2

Answer:

$\frac{1}{4}F$ is the force on the two given charge .

Explanation:

Given ,   two charges of same magnitude but of opposite nature are placed by some distance away .

Force F netwon is acting between the given two charge and

Let $q_1 and \ q_2$ be two charge .

As we know Force (F) = $\frac{kq_1q_2}{r^2}$   where k is constant and r is the distance between them .

Now according to the question 50 % of one charge is transfered to another charge .

⇒$q_1$ = $q - \frac{q}{2}$ = $\frac{q}{2}$

and $q_2$ = $-q + \frac{q}{2}$ = $\frac{-q}{2} = \frac{q}{2}$

Therefore , force on them is ;

$F'$ = $\frac{kq_1 q_2}{r^2}$  ..........(i)

Now put the value of $q_1 and\ q_2$ in (i) we get ,

$F'$ = $\frac{k\frac{q}{2} \frac{q}{2} }{r^2}$ = $\frac{1}{4} \frac{kq^2}{r^2}$= $\frac{1}{4}F$        [Where F = $\frac{kq^2}{r^2}$]

Hence , the force on them is $\frac{1}{4}F$ .

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