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Electric Potential - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Electric potential is considered one of the most asked concept.

  • 48 Questions around this concept.

Solve by difficulty

Two thin wire rings each having a radius R are placed at a distance apart with their axes coinciding. The charges on the two rings are +Q and -Q  The potential difference between the centers of the two rings is

An electric charge 10^{-3}\mu C  is placed at the origin (0,0)of  X-Y co-ordinate system . Two points A and B are situated at \left ( \sqrt{2},\sqrt{2} \right )\: and\: \left ( 2,0 \right ) respectively. The potential difference between the point  A and B will be

A charge Q is uniformly distributed over a long rod AB of length L, as shown in the figure. The electric potential at the point O lying at a distance L from the end A is :

 

Two points P\; and\; Q are maintained at the potentials of 10 V and -4 V respectively. The work done in moving 100 electrons from P\; to\; Q  is

This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement-1:  For a charged particle moving from point P to point Q , the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q.

Statement-2:  The net work done by a conservative force on an object moving along a closed loop is zero.

 

Concepts Covered - 1

Electric potential

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_{ext} =\int {\overrightarrow{F_{ext}}\cdot \overrightarrow{dr}}

Since

W_{ext}=\Delta U \ \ and \ \ \Delta {KE}=0 \\ \Rightarrow \text{For all time }\ \ F_{net}=0 \\ \Rightarrow \overrightarrow{F}_{ext} +\overrightarrow{F}_{system}=0

i.e  \overrightarrow{F}_{ext} = -\overrightarrow{F}_{system}     
  

So V=\frac{W_{ext}}{q_0}=\int \frac{\overrightarrow{F}_{ext}\cdot \overrightarrow{dr}}{q_0}=-\int \frac{\overrightarrow{F}_{system}\cdot \overrightarrow{dr}}{q_0}

where

V\rightarrow Electric potential

  •  It is a scalar quantity.

  •  SI \ \ Unit \to \;\frac{J}{C}=volt while  CGS unit is stat volt  

           1 \ volt=\frac{1}{300} \ stat \ volt.   

  • Dimension -   \begin{aligned}[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] \end{aligned}.

   

  • Electric Potential at a distance 'r'

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

F= \frac{Kqq_0}{r^{2}}

Using V=\frac{W_{ext}}{q_0}=\int \frac{\overrightarrow{F}_{ext}\cdot \overrightarrow{dr}}{q_0}=-\int \frac{\overrightarrow{F}_{system}\cdot \overrightarrow{dr}}{q_0}

          V=\frac{Kq}{r}

                at r=\infty       V=0=V_{max}

  • 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.

V_{A}=Electric potential at point A

V_{B}=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.

            \Delta V=V_{B}-V_{A}=\frac{W_{A\rightarrow B}}{q}

           \Delta V=KQ\left [ \frac{1}{r_{B}}-\frac{1}{r_{A}} \right ]

  • 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 (Q1,Q2,Q3…) can be calculated by doing a scalar sum of their individuals Electric potential.

 V=V_{1}+V_{2}+V_{3}+\cdots=\frac{kQ_{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

                  \dpi{100} V=\int dV=\int \frac{dq}{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

 \left ( It \: is \: assumed\: that\left | Q_{1} \right |< \left | Q_{2} \right | \right )

Let at P, V is zero

 \\*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 )}

 

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

\\*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 )}          

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Electric potential

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