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Relation Between Electric Field And Potential MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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  • Relation between electric field and potential is considered one of the most asked concept.

  • 32 Questions around this concept.

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Assume that an electric field \vec{E} = 30 x^{2} \hat{i} exists in space.  Then the potential difference VA - VO, where VO is the potential at the origin and VA the potential at x=2 m is :

 An electric field is given byE _x = - 2 x^3 kN/C. The potential of the point (1, –2), if potential of the point (2, 4) is taken as zero, is 

The electric potential V at any point (x, y, z), all in meters in space is given by V = 4x2 volt. The electric field at the point (1, 0, 2) in volt/meter is 

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Relation between electric field and potential

Electric field and potential are related as

\vec{E}=-\frac{dV}{dr}

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}{dx},E_y=\frac{\delta V}{dy},E_z=\frac{\delta V}{dz}

where

E_{x}=-\frac{\partial V}{dx}  (partial derivative of V w.r.t. x)

E_{y}=-\frac{\partial V}{dy}    (partial derivative of V w.r.t. y)

E_{z}=-\frac{\partial V}{dz}    (partial derivative of V w.r.t. z)

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

 

dW=\vec{F}.\vec{r}=q\vec{E}.d\vec{r}

Electric potential V is defined as negative of work done per unit charge

 dV=-\frac{dW}{q}

So  Integrating between r1, and r2

We get V(\vec{r_2})-V(\vec{r_1})= \int_{r_1}^{r_2} -\vec{E}.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}.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}=dx \ \vec{i}+dy \ \vec{j}+dz \ \vec{k}

So  

\vec{E}.d\vec{r}=-dV=E_xdx +E_ydy +E_zdz\\ dV=-E_xdx -E_ydy -E_zdz

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

Thus E_x=\frac{dV}{dx}

Similarly E_y=\frac{dV}{dy}, E_z=\frac{dV}{dz} 

 

  • When an electric field is a uniform (constant)

As Electric field and potential are related as dV = \int_{r_0}^{r} -\vec{E}.d\vec{r}

and E=constant then  dV =-\vec{E}\int_{r_0}^{r} d\vec{r}=-\vec{E}dr

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

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Relation between electric field and potential

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