A two point charges 4 q and – q are fixed on the x -axis at x = − d 2 and x = d 2 , respectively. If the third point charge ‘ q ’ is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will:

Step-by-step solution

Potential at origin O , the potential due to two charges is given by V 0 = 4 q 4 π ε 0 d 2 + − q 4 π ε 0 d 2 = 6 q 4 π ε 0 d Potential at P at a distance, x = d is given by V P = 4 q 4 π ε 0 3 d 2 + − q 4 π ε 0 d 2 = q 6 π ε 0 d Change in potential energy due the charge motion of charge q is given by ∆ U = q Δ V = q V f − V i ⇒ ∆ U = q V P − V O ∆ U = q q 6 π ε 0 d − 3 q 2 π ε 0