Electric Potential - Practice Questions & MCQ

In an Electric field Electric potential V at a point, P is defined as work done per unit charge in changing the position of test charge from some reference point to the given point.

Note-usually reference point is taken as infinity and potential at infinity is taken as Zero.

We know that

$
W_{e x t}=\int \overrightarrow{F_{e x t}} \cdot \overrightarrow{d r}
$

Since

$
W_{\text {ext }}=\Delta U \text { and } \Delta K E=0
$

$\Rightarrow \underset{\text { For all time }}{ } F_{n e t}=0$
$\Rightarrow \vec{F}_{\text {ext }}+\vec{F}_{\text {system }}=0$
i.e $\vec{F}_{\text {ext }}=-\vec{F}_{\text {system }}$

So $V=\frac{W_{\text {ext }}}{q_0}=\int \frac{\vec{F}_{\text {ext }} \cdot \overrightarrow{d r}}{q_0}=-\int \frac{\vec{F}_{\text {system }} \cdot \overrightarrow{d r}}{q_0}$
where
$V \rightarrow$ Electric potential
- It is a scalar quantity.

SI Unit $\rightarrow \frac{J}{C}=$ volt ${ }_{\text {while CGS unit is stat volt }}$
1 volt $=\frac{1}{300}$ stat volt.
- Dimension -

$
[V]=\left[\frac{W}{q_0}\right]=\left[\frac{M L^2 T^{-2}}{A T}\right]=\left[M L^2 T^{-3} A^{-1}\right]
$
 

• Electric Potential at a distance 'r'

            If the Electric field is produced by a point charge q then 

$\begin{aligned} & F=\frac{K q q_0}{r^2} \\ & \qquad \begin{array}{l}V=\frac{W_{\text {ext }}}{q_0}=\int \frac{\vec{F}_{\text {ext }} \cdot \overrightarrow{d r}}{q_0}=-\int \frac{\vec{F}_{\text {system }} \cdot \overrightarrow{d r}}{q_0} \\ \qquad V=\frac{K q}{r} \\ \quad \text { at } r=\infty \quad V=0=V_{\max }\end{array}\end{aligned}$

• Electric Potential difference

In the Electric field, the work done to move a unit charge from one position to the other is known as Electric Potential difference.

If the point charge Q is producing the field

Point A and B are shown in the figure.

=Electric potential at point A

=Electric potential at point B

$r_B \rightarrow$ the distance of charge at $B$
$r_A \rightarrow$ distance of charge at $A$
$\Delta V=$ The Electric potential difference in bringing charge q from point A to point B in the Electric field produced by Q .

$
\begin{aligned}
& \Delta V=V_B-V_A=\frac{W_{A \rightarrow B}}{q} \\
& \Delta V=K Q\left[\frac{1}{r_B}-\frac{1}{r_A}\right]
\end{aligned}
$

- Superposition of Electric potential

Statement- Total electric potential at a given point in space due to all the charges placed around it is the scalar or algebraic addition of electric potential due to individual charges at that point.
i.e

The net Electric potential at a given point due to different point masses $\left(Q_1, Q_2, Q_3 \ldots\right)$ can be calculated by doing a scalar sum of their individuals Electric potential.

$
V=V_1+V_2+V_3+\cdots=\frac{k Q_1}{r_1}+k \frac{Q_2}{r_2}+\frac{k\left(Q_3\right)}{r_3}+\cdots=\frac{1}{4 \pi \varepsilon_0} \sum \frac{Q_i}{r_i}
$
 

• - Electric Potential due to Continuous charge distribution

  $
  V=\int d V=\int \frac{d q}{4 \pi \varepsilon_0 r}
  $
   

• Graphical representation

     As we move on the line joining two charges then the variation of Potential with distance is given below.

• Zero potential due to a system of a two-point charge

1.For internal point
(It is assumed that $\left|Q_1\right|<\left|Q_2\right|$ )
Let at $P, V$ is zero

$
\begin{aligned}
& V_P=0 \Rightarrow \frac{Q_1}{x_1}=\frac{Q_2}{\left(x-x_1\right)} \\
& \Rightarrow x_1=\frac{x}{\left(Q_2 / Q_1+1\right)}
\end{aligned}
$
 

If both charges are like then the resultant potential is not zero at any finite point.

2.For external point

Let at P, V is zero

$\begin{aligned} & V_P \Rightarrow \frac{Q_1}{x_1}=\frac{Q_2}{\left(x+x_1\right)} \\ & \Rightarrow x_1=\frac{x}{\left(Q_2 / Q_1-1\right)}\end{aligned}$