# Electric field vector in a region is given by $$\vec{E}=(3\hat{i}+4y\hat{j}) V/m$$. The potential at the origin is zero. Then the potential at a point (2,1) m is

a) 7 V

b) 8 V

c) -8 V

d) -7 V

Solution:

$$\begin{array}{l}\vec{E}=(3\hat{i}+4y\hat{j})\end{array}$$

Ex = 3

Ey = 4y

$$\begin{array}{l}V= -\int_{0}^{2}E_{x}dx-\int_{0}^{1}E_{y}dy\end{array}$$

$$\begin{array}{l}V= -\int_{0}^{2}3dx-\int_{0}^{1}4ydy\end{array}$$

$$\begin{array}{l}= -3[x]_{0}^{2}-2[y^{2}]_{0}^{1}\end{array}$$

= – 3 x 2 – 2 x 1

= – 8 V