Question

E=20i+30j exist in space.If the potential at origin is taken to be zero,what is the potential at p(3,2)

Answer 1

\(V= - \int \vec E d \vec l=-|\vec E||\vec l| \cos \theta\), where E is a constant.
taking from point(0,0) to (3,2)
\(|\vec E|=\sqrt {20^2 +30^2}=10\sqrt{13}\), where the angle is \(\tan^{-1}\frac{3}{2}\)
now for \(|\vec l|= \sqrt {2^2 +3^2}=\sqrt{13}\), where angle is \(\tan^{-1}\frac{2}{3}\)
\(\cos \theta=\tan^{-1}\frac{3}{2}-\tan^{-1}\frac{2}{3}\)
sub it all in.

Answer 2

cosθ=tan−13/2−tan−12/3. I can not understand it.

Answer 3

the angle between the field vector and the displacement vector is \(tan^{−1}3/2−tan^{−1}2/3\)