Question

Answer this Question of iit jee .
Chapter → Electrostatics​​

Answer 1

EXPLANATION.

Two charges each equal to q.

Kept x = - a and x = a on the x-axes.

Particle of mass m and charge $q_{o}$ = q/2 is placed at origin.

$q_{o}$ is a small displacement (y <<a) along the y-axes.

As we know that,

$\implies F = \dfrac{kq^{2} }{2(y^{2} + a^{2} )}$

$\implies F_{V} = F cos \theta + F cos \theta = 2F cos \theta$

$\implies F_{H} = F sin \theta - F sin \theta = 0.$

$\implies F_{net} = 2F cos \theta$

$\implies F_{net} = \dfrac{2kq^{2} }{2(y^{2} + a^{2}) }cos \theta \ = \dfrac{kq^{2} }{(y^{2} + a^{2} )} cos \theta$

$\implies F_{net} = \dfrac{kq^{2} }{(y^{2} + a^{2} )} \times \dfrac{y}{\sqrt{y^{2} + a^{2} } }$

$\implies F_{net} = \dfrac{kq^{2}y }{(y^{2} + a^{2} )(y^{2} + a^{2} )^{1/2} }$

$\implies F_{net} = \dfrac{kq^{2}y }{(y^{2} + a^{2})^{3/2} }$

In the question it is already given that (y <<a), it means.

y² + a² ≈ a².

Substitute the value in the equation, we get.

$\implies F_{net} = \dfrac{kq^{2}y }{(a^{2})^{3/2} } \ = \dfrac{kq^{2}y}{a^{3} }$

F ∝ y.

Option [D] is correct answer.

Answer 2

plz mark me as brainlist