Question

plz give me the answer of this question ​

Answer 1

Answer:

Electric field due to p1 at a point on r distance on axis

$= \frac{ 2kp_{1}}{ {r}^{3} }$

Electric potential energy of p2 due to this electric field

$U = - p_2( \frac{2k \:p_1}{ {r}^{3} })$

So mutual force

$f = \frac{ - dU}{dr} \\ \\ = - \frac{d}{dr} (\frac{ 2kp_{1}p_{2}}{ {r}^{3} }) \\ \\ = - 2kp_{1}p_{2} \: ( \frac{ - 3}{ {r}^{4} } ) \\ \\ = \frac{6kp_{1}p_{2}}{ {r}^{4} }$