The potential at a point x (measured in $\mu$m) due to some charges situated on the x-axis is given by $V(x)=\dfrac{20}{x^{2}-4}$ volts. The electric field E at $x = 4\ \mu$m is given by

(D) $\dfrac{10}{9}$ V/$\mu$m and in the $+x$ direction

(A) $\dfrac{5}{3}$ V/$\mu$m and in the $-x$ direction

(B) $\dfrac{5}{3}$ V/$\mu$m and in the $+x$ direction

(C) $\dfrac{10}{9}$ V/$\mu$m and in the $-x$ direction

JEE Main · 2007 · 29 Apr 2007 · Offline · Electrostatics

The potential at a point x (measured in

μ

m) due to some charges situated on the x-axis is given by

V (x) = \frac{20}{x ^{2} - 4}

volts. The electric field E at

x = 4 μ

m is given by

Answer: (D)

The field is minus the gradient of the potential,

E = - \frac{d V}{d x}

, with

V = \frac{20}{x ^{2} - 4}

. Differentiating,

E = - 20 \cdot \frac{- 2 x}{( x ^{2} - 4 ) ^{2}} = \frac{40 x}{( x ^{2} - 4 ) ^{2}}

. At

x = 4

this is

\frac{160}{144} = \frac{10}{9}

volt per micrometre, and because the value comes out positive the field points along the direction of increasing

x

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option