Question

Hey guys plz answer this question ​

Answer 1

Given :

Two point charges q ,2q are placed at (a,0) and (0,a) respectively on shown .

Electrostatic Force on Charge Q placed at origin is :

Theory :

Coulombs law :

It States that two point charges at rest separated by distance r exert a force on on each other ,which is directly proportional to the product of the magnitude of charges,and inversely proportional to square of the distance between them.

$f = \dfrac{kq_{1}q_{2} }{r {}^{2} }$

Solution :

Force exert by q charge on Q

$= \dfrac{k \times q \times Q}{a {}^{2} }$

$= f$

Force exert by 2q charge on Q

$= \dfrac{k \times 2q \times Q}{a {}^{2} }$

$= 2f$

Now resultant on 2f and f forces

$f_{net} = \sqrt{f {}^{2} + (2f) {}^{2} } = \sqrt{5} f$

Therefore ;Electrostatic Force on Charge Q placed at origin is is √5f

$f _{net} = \sqrt{5 } \times \dfrac{kqQ}{a{}^{2} }$

Correct option b)

For step by step explanation refer to the attachment.