Question

A uniform electric field having strenght is existing in x-y plane as shown in figure. Find the p.d. between origin O & A (d,d,0),

• A. $Ed(cos\theta +sin\theta )$
• B. $-Ed(sin\theta -cos\theta )$
• C. $\sqrt{2}Ed$
• D. None of these

Answer 1

The correct option is A $Ed(cos\theta +sin\theta )$

When the electric field E in x-y plane makes angle $\theta$ with the x-axis , the electric filed can be written as $\overrightarrow{E}=E\cos \theta \hat{i}+E\sin \theta \hat{j}$

Thus, $E_x=E\cos \theta$ , $E_y=E\sin \theta$ and $E_z=0$

Since $\overrightarrow{E}=-\frac{\overrightarrow{\nabla }V}{dr}$

So, $d{V}_x=-E_{x}dx$, $d{V}_y=-E_{y}dy$ and $d{V}_z=-E_{z}dz$

Now ingratiating, $V_x=-{\int }_d^{0}E\cos \theta dx=-E\cos \theta (0-d)=Ed\cos \theta$

$V_y=-{\int }_d^{0}E\sin \theta dy=-E\sin \theta (0-d)=Ed\sin \theta$

and $V_z=0$

Thus, potential between O and A is $V=V_x+V_y+V_z=Ed\cos \theta +Ed\sin \theta$