Relation Between Electric Field And Potential MCQ - Practice Questions with Answers

Electric field and potential are related as

$\overrightarrow{E}=-\frac{dV}{dr}$

Where E is Electric field
And V is Electric potential
And $r$ is the position vector
And Negative sign indicates that in the direction of intensity the potential decreases.
If $\overrightarrow{r}=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$

Then

$E_x=\frac{\delta V}{dx},E_y=\frac{\delta V}{dy},E_z=\frac{\delta V}{dz}$

where

$\begin{array}{rl}& E_x=-\frac{\partial V}{dx}(\text{ partial derivative of }\mathrm{V}\text{ w.r.t. x) }\\ & E_y=-\frac{\partial V}{dy}\text{ (partial derivative of }\mathrm{V}\text{ w.r.t. y) }\\ & E_z=-\frac{\partial V}{dz}\text{ (partial derivative of }\mathrm{V}\text{ w.r.t. z) }\end{array}$
 

Proof-

Let the Electric field at a point r due to a given mass distribution is E.

If a test charge q is placed inside a uniform Electric field E.

Then force on a charged particle q when it is at r is  $\overrightarrow{F}=q\overrightarrow{E}$ as shown in figure

      

 

As the particle is displaced from r to r + dr the

work done by the Electric force on it is

 

$dW=\overrightarrow{F}\cdot \overrightarrow{r}=q\overrightarrow{E}\cdot d\overrightarrow{r}$

Electric potential V is defined as negative of work done per unit charge

$dV=-\frac{dW}{q}$

So Integrating between $r_1$, and $r_2$

We get

$V\left(\overrightarrow{r_2}\right)-V\left(\overrightarrow{r_1}\right)={\int }_{r_1}^{r_2}-\overrightarrow{E}\cdot d\overrightarrow{r}$

If $r_1=r_0$, is taken at the reference point, $V\left(r_0\right)=0$.
Then the potential $V(r_2=r)$ at any point $r$ is

$V(\overrightarrow{r})={\int }_{r_0}^r-\overrightarrow{E}\cdot d\overrightarrow{r}$

in Cartesian coordinates, we can write

$\overrightarrow{E}=E_x\overrightarrow{i}+E_y\overrightarrow{j}+E_z\overrightarrow{k}$
 

If $\overrightarrow{r}=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$
Then $d\overrightarrow{r}=dx\overrightarrow{i}+dy\overrightarrow{j}+dz\overrightarrow{k}$
So

$\begin{array}{c}\overrightarrow{E}\cdot d\overrightarrow{r}=-dV=E_{x}dx+E_{y}dy+E_{z}dz\\ dV=-E_{x}dx-E_{y}dy-E_{z}dz\end{array}$

If y and z remain constant, $\mathrm{dy}=\mathrm{dz}=0$
Thus $E_x=\frac{dV}{dx}$

$E_y=\frac{dV}{dy},E_z=\frac{dV}{dz}$

- When an electric field is a uniform (constant)

As Electric field and potential are related as

$dV={\int }_{r_0}^r-\overrightarrow{E}\cdot d\overrightarrow{r}$
 

and E=constant then  $dV=-\overrightarrow{E}{\int }_{r_0}^{r}d\overrightarrow{r}=-\overrightarrow{E}dr$

• If at any region E = 0 then V = constant
• If V = 0 then E may or may not be zero.