Question

An electric field vector $\vec{E}= (25 \hat{i}+ 30 \hat{j}) N{C}^{-1}$ exists in a region of space. If the potential at the origin is taken to be zero then the potential at $x=2m, y=2m$ is
A. -110J
B. -140J
C. -130J
D. -120J

Answer 1

To solve this problem, use the relationship between electric field vector, electric potential and displacement vector. Find the displacement vector using the x and y coordinates mentioned in the question. Substitute the values of the electric field vector and displacement vector in the above-mentioned relationship and find the electric potential. This will be the obtained potential at $x= 2m$ and $y=2m$.

Complete step-by-step solution:
$\vec{E}= (25 \hat{i}+ 30 \hat{j})$ ...(1)
We have to find the potential at x=2m and y=2m
$\Rightarrow \vec{r}= (2\hat{i}+ 2 \hat{j})$
We know, relationship between electric field and electric potential is given by,
$V = -\vec {E}. \vec {r}$
Where, V is the electric potential
$\vec{E}$ is the electric field
$\vec{r}$ is the displacement vector
Substituting values in above equation we get,
$V=- (25 \hat{i}+ 30 \hat{j})(2 \hat{i}+ 2 \hat{j})$
$\Rightarrow V=-[ (25 \times 2)(\hat{i}\times \hat{i})+ (25 \times 2)(\hat{i}\times \hat{j})+ (30 \times 2)(\hat{j}\times \hat{i})+ (30 \times 2)(\hat{j}\times \hat{j})]$
$\Rightarrow V= -[50 \times 1 + 40 \times 0+ 60 \times 0 + 60 \times 1]$
$\Rightarrow V= -50 - 0 – 0 – 60$
$\Rightarrow V= -110 J$
Thus, the potential at $x=2m, y=2m$ is $-110J.$
So, the correct answer is option A i.e. $-110 J$.

Note: Electric field is the negative gradient of potential which depends inversely upon the distance of a given point of interest from a charge. The direction in which the unit positive charge moves shows the direction of the electric field. Thus, the electric field is a vector quantity. The direction of the electric field is that in which the potential decreases sharply. Students must take care of the negative sign in the relationship between electric field vectors and electric potential.