Relation Between Electric Field And Potential - Practice Questions & MCQ

Electric field and potential are related as

$
\vec{E}=-\frac{d V}{d r}
$

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 $\vec{r}=x \vec{i}+y \vec{j}+z \vec{k}$

Then $E_x=\frac{\delta V}{d x}, E_y=\frac{\delta V}{d y}, E_z=\frac{\delta V}{d z}$
where

$
\begin{aligned}
E_x & =-\frac{\partial V}{d x} \quad \text { (partial derivative of } \mathrm{V} \text { w.r.t. x) } \\
E_y & =-\frac{\partial V}{d y} \quad \text { (partial derivative of } \mathrm{V} \text { w.r.t. y) } \\
E_z & =-\frac{\partial V}{d z} \quad \text { (partial derivative of } \mathrm{V} \text { w.r.t. z) }
\end{aligned}
$

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 $\vec{F}=q \vec{E}$ as shown in figure

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

work done by the Electric force on it is

$
d W=\vec{F} \cdot \vec{r}=q \vec{E} \cdot d \vec{r}
$

Electric potential V is defined as the negative of work done by electric force per unit charge

$
d V=-\frac{d W}{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}-\vec{E} \cdot d \vec{r}
$

If $r_1=r_0$, is taken at the reference point, $\mathrm{V}\left(\mathrm{r}_0\right)=0$.
Then the potential $V\left(r_2=r\right)$ at any point $r$ is

$
V(\vec{r})=\int_{r_0}^r-\vec{E} \cdot d \vec{r}
$

in Cartesian coordinates, we can write

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

If $\vec{r}=x \vec{i}+y \vec{j}+z \vec{k}$
Then $d \vec{r}=d x \vec{i}+d y \vec{j}+d z \vec{k}$
So

$
\begin{gathered}
\vec{E} \cdot d \vec{r}=-d V=E_x d x+E_y d y+E_z d z \\
d V=-E_x d x-E_y d y-E_z d z
\end{gathered}
$

If y and z remain constant, $\mathrm{dy}=\mathrm{dz}=0$

$
{ }_{\text {Thus }} E_x=\frac{d V}{d x}
$
 

Similarly

$
E_y=\frac{d V}{d y}, E_z=\frac{d V}{d z}
$

- When an electric field is a uniform (constant)

As Electric field and potential are related as

$
d V=\int_{r_0}^r-\vec{E} \cdot d \vec{r}
$

and $\mathrm{E}=$ constant then

$
d V=-\vec{E} \int_{r_0}^r d \vec{r}=-\vec{E} d r
$

- If at any region $\mathrm{E}=0$ then $\mathrm{V}=$ constant
- If $\mathrm{V}=0$ then E may or may not be zero.