Question

The electric field $\overrightarrow{E}=(20\hat{i}+30\hat{j})\text{N/C}$ exists in space. If the potential at origin in taken to be zero, find the potential at $P(3,2)$.

• A. $150\text{V}$
• B. $-150\text{V}$
• C. $-120\text{V}$
• D. $120\text{V}$

Answer 1

The correct option is C $-120\text{V}$

The relation between electric field $\left(\overrightarrow{E}\right)$ and electric potential $\left(V\right)$ is
$V=-\int \overrightarrow{E}.\overrightarrow{ds}\dots \left(i\right)$
Where $\overrightarrow{ds}$ is distance in direction of electric field.
Let $dx$ and $dy$ are elemental distance in $x$ and $y$ direction.
So,
$\overrightarrow{ds}=dx\hat{i}+dy\hat{j}$
Given $\overrightarrow{E}=(20\hat{i}+30\hat{j})\text{N/C}$

Substituting in equation $\left(i\right)$
$\Rightarrow V_P=-\underset{(0,0)}{\overset{(3,2)}{\int }}(20\hat{i}+30\hat{j}).(dx\hat{i}+dy\hat{j})$
$\Rightarrow V_P=-\underset{(0,0)}{\overset{(3,2)}{\int }}(20dx\hat{i}+30dy\hat{j})$
$\Rightarrow V_P=-(20x{|}_0^3+30y{|}_0^2)$
$\Rightarrow V_P=-(20\times 3+30\times 2)$
$\Rightarrow V_P=-120\text{V}$

Hence, option (b) is correct.

$\begin{array}{c}\text{Why this question ?}\\ \text{Tip: If electric field is given in vector}\\ \text{form then the potential in x, y, z}\\ \text{direction can be found and then added.}\end{array}$