Question

The potential at a point x (measured in μm) due to some charges situated on the x-axis is given by V(x) = [(20)/(x² – 4)] Volt. The electric field at x = 4μm is given by 

(A) (5/3) Vμm^–1 and in positive x - direction 

(B) (10/9) Vμm^–1 and in positive x - direction 

(C) (10/9) Vμm^–1 and in positive x - direction 

(D) (5/3) Vμm^–1 and in positive x – direction

Answer 1

The correct option (B) (10/9) Vμm^–1 and in positive x – direction

Explanation:

V(x) = [(20)/(x² – 4)]

Electric field = E(x) = – (dV/dx)

∴ E(x) = – 20 (d/dx) [1/(x² – 4)]

E(x) = (– 20) × [(– 1)/(x² – 4)²] (2x – 0) = [(40x)/(x² – 4)²]

hence E(x) at x = 4μm is

E(x) = [(40 × 4)/(4² – 4)²] = [(160)/(144)] (10/9) V/μm and it is in positive x direction