Q. A uniform electric field having a magnitude $E_{0}\ $and direction along the positive X –axis, exists. If the potential V is zero at $x = 0$, then its value at $x = + x$ will be
A) $V(x) = + x{\ E}_{0}$
B) $V(x) = - x{\ \ E}_{0}$
C) $V(x) = + x^{2}{\ E}_{0}$
D) $V(x) = - x^{2}{\ E}_{0}$
Explanation
(b) As $\ \ \ \ E = - \frac{dV}{dr}$
$\therefore\ \ + E_{0} = - \frac{\left\lbrack V(x) - 0 \right\rbrack}{x}$
or $V_{x} = - E_{0}x$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.