Relation Between Electric Field And Potential

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.