Relation Between Electric Field And Potential

Imagine yourself climbing a hill and the effort it takes to move upwards; similarly, in electricity, charges feel this effort or force during their movements in the presence of an electric field. That relation of electric field-containing force, an electric charge experiences with potential energy available at different points, now electric potential, is the reason that backs the functioning of electric circuits and fields. Let us look at the relationship between electric fields and potential, how they interact, and why this understanding is important.

The topic of electric field and potential is important, and in this article, we steady about it. This concept we study in the chapter electrostatics which is a crucial chapter in Class 12th physics and it is a high-wage chapter. It is not only essential for board exams but also for competitive exams like the JEE, NEET, and other entrance exams such as SRMJEE, BITSAT, WBJEE, VITEEE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of eight questions have been asked on this concept. And for NEET two questions were asked from this concept.

Relation Between Electric Field and Potential

Electric field and potential are related as

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

Where E is the Electric field, V is the Electric potential and r is the position vector.

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

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

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 $\mathrm{q}$ when it is at $\mathrm{r}$ 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 r₁, and r₂

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 r₁=r₀ is taken at the reference point, V(r₀) = 0.

Then the potential V(r₂=r) at any point r is

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

in Cartesian coordinates, we can write

$\begin{aligned}
& \vec{E}=E_x \vec{i}+E_y \vec{j}+E_z \vec{k} \\
& \text { If } \vec{r}=x \vec{i}+y \vec{j}+z \vec{k}
\end{aligned}$

Then,

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

So

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

If y and z remain constant, dy = dz = 0

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 uniform (constant)

The electric field and potential are related as $\quad d V=\int_{r_0}^r-\vec{E} \cdot d \vec{r}$
and $\mathrm{E}=$ constant then
$
d V=-\vec{E} \int_{r_0}^r d \vec{r}=-\vec{E} d r
$

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

For More Information On Relation Between Electric Field And Potential, Watch The Below Video:

Solved Example Based On Relation Between Electric Field And Potential

Example 1: The potential at a point $x$ (measured in $\mu m$ ) due to some charges situated on the $x$-axis is given by:
$$
V(x)=20 /\left(x^2-4\right) \text { volt } .
$$

$\text { The electric field } E \text { at } x=4 \mu m \text { is given by }$

1) (10/9)volt/ $\mu \mathrm{m}$ and in the + ve $x$ direction
2) $(5 / 3)$ volt/ $\mu \mathrm{m}$ and in the $-v e x$ direction
3) $(5 / 3)$ volt $/ \mu \mathrm{m}$ and in the + ve $x$ direction
4) (10/9)volt/ $\mu \mathrm{m}$ in the - ve $x$ direction

Solution:

As we learnt in

Relation between field and potential -

$
E=\frac{-d V}{d r}
$
- wherein
$\frac{d V}{d r}-\underset{\text { Potential gradient. }}{ }$
Electric field $E=\frac{-d V}{d x}=\frac{-d}{d x}\left(\frac{20}{x^2-4}\right)=\frac{40 x}{\left(x^2-4\right)^2}$
At $x=4 \mu m$
$
\therefore \quad E=\frac{40 \times 4}{[16-4]^2}=\frac{160}{144}=\frac{10}{9} \mathrm{~V} / \mu \mathrm{m}
$

Positive signs indicate E is in the +ve x direction

Hence, the answer is option (1).

Example 2: Two charges +q and -q are situated at a certain distance. At the point exactly midway between them

1)Electric field and potential both are zero

2)The electric field is zero but the potential is not zero

3)The electric field is not zero but the potential is zero

4)Neither electric field nor potential is zero

Solution:

If at any point $E=0$ -
$\mathrm{V}=$ constant
At $O, E \neq 0, V=0$

Hence, the answer is option (3).

Example 3: A uniform electric field having a magnitude $E_0$ and direction along the positive $X$-axis exists. If the electric potential $V$, is zero at $X=0$, then, its value at $X=+x$ will be:

1) $V(X)=+X E_0$
2) $V(X)=-X E_0$
3) $V(X)=X^2 E_0$
4) ${ }_{} V(X)=-X^2 E_0$

Solution:

As we learned

In space -

$
E_x=\frac{-d v}{d x} \quad E_y=\frac{-d v}{d y} \quad E_z=\frac{-d v}{d z}
$

By using
$
E=-\frac{\Delta V}{\Delta r}=-\frac{\left(V_2-V_1\right)}{\left(r_2-r_1\right)} ; E_0=\frac{\{V(x)-0\}}{x-0} \Rightarrow V(x)=-x E_0
$

Hence, the answer is option (2).

Example 4: If the electric potential at any point $(\mathrm{x}, \mathrm{y}, \mathrm{z}) \mathrm{m}_{\text {in }}$ space is given by $\mathrm{V}=3 \mathrm{x}^2$ volt. |The electric field at the point $(1,0,3) \mathrm{m}$ will be :

1) $3 \mathrm{Vm}^{-1}$, directed along positive $x$-axis,
2) $3 \mathrm{Vm}^{-1}$, directed along negative $x$-axis.
3) $6 \mathrm{Vm}^{-1}$, directed along positive $x$-axis.
4) $6 \mathrm{Vm}^{-1}$, directed along negative $x$-axis.

Solution:

$\begin{aligned}
& \mathrm{V}=3 \mathrm{x}^2 \\
& \mathrm{E}=\frac{-\mathrm{dv}}{\mathrm{dx}}=-6 \mathrm{x} \\
& \vec{E}=-6 x \hat{\imath} \\
& \text { at }(1,0,3) \\
& \vec{E}=-6 \hat{\imath}
\end{aligned}$

Hence, the answer is option (4).

Example 5: Some equipotential surfaces are shown in the figure. The magnitude and direction of the electric field is

1) 100 V/m making angle 120^o with the x-axis

2) 100 V/m making angle 60^o with the x-axis

3) 200 V/m making angle 120^o with the x-axis

4) None of the above

Solution:

As we learned

For Positive charge -

An electric field line comes out.

By using $d V=E d r \cos \theta$ suppose we consider line 1 and line 2 then
$
(30-20)=E \cos 60^{\circ}(20-10) \times 10^{-2}
$

So
$E=200 \mathrm{volt} / \mathrm{m}$ making in angle $120^{\circ}$ with $x$-axis

Hence, the answer is option (3).

Summary

That is, the electric field will be the force per unit charge experienced by a charged particle in a field; therefore, electric potential is usually expressed as the energy per unit charge at a point in the field. That is, the electric field will be related to an electric potential given by the relation of a field pointing toward a maximum decrease in potential. It is mathematically the negative gradient of electric potential. This enables us to study and design electrical systems and learn about the behaviour of charges in fields.