Answer to Question #134302 in Electricity and Magnetism for Rohan

Question

The electric field in a region is given by Ē = 20î + 30ĵ in certain units. As one goes a small distance dl in the direction of ñ = 30î - 20ĵ , the potential will :-

(a) increase

(b) decrease

(c) remain constant

(d) change sign

(Note:- This question have one or more than one correct choice(s) out of the four given choices. Any number of options may be correct.)

Answer 1

We know that $\vec{E} = -\nabla \varphi.$

The change of potential is $\varphi(A) - \varphi(B) = \int\limits_A^B \vec{E}\cdot d\vec{l}$ .

For the small distance we'll get $\varphi(A) - \varphi(B) \approx \vec{E}\cdot d\vec{n} = (20i+30j)(30i-20j) = 60 - 0 + 0 - 60 = 0.$

So the change of potential is zero. It is clear, because the change is perpendicular to the electric field, so the potential remains constant.