Tag: coulomb's law

Questions Related to coulomb's law

The electric potential decreases uniformly from 120 V to 80 V as one moves on the x-axis from $x = -1\ cm$ to $ x = +1 \ cm$. The electric field at the origin.

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. must be equal to 20 Vcm$^{-1}$

  2. must be equal to 20 Vm$^{-1}$

  3. greater than or equal to 20 Vcm$^{-1}$

  4. may be less than 20 Vcm$^{-1}$

Show answer
Correct Option: A,C
Explanation:

$dv=-\vec E.\vec dx=-E dx \cos\theta$

$\displaystyle E = -\dfrac{dV}{\cos\theta dx} $

$\cos\theta\approx 1$, if we take $\cos\theta =1$,then

$E _{min}= -\dfrac{80-120}{1-(-1)} =\dfrac{40}{2}= 20 V cm^{-1}$

Hence E would be greater than or equal to $20$

Mark the correct statement:

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. If $E$ is zero at a certain point, then $V$ should be zero at that point

  2. If $E$ is not zero at a certain point, then $V$ should not be zero at that point

  3. If $V$ is zero at a certain point, then $E$ should be zero at that point

  4. If $V$ is zero at a certain point, then $E$ may or maynot be zero

Show answer
Correct Option: D
Explanation:

Since $E=\dfrac{-dV}{dr}$

A zero potential at a point never means that the electric field is also zero at a point always.
$E=0$ when $V=$ constant.
If $V=0$; then $E$ may or may not be zero.
Similarly, if $E=0$; then $V$ must either be a constant or may be zero.
so correct option is(d).

For a uniform electric field $\vec{E}=E _{0}(\hat{i})$, if the electric potential at x=0 is zero, then the value of electric potential at x=+x will be .......

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $xE _{0}$

  2. -$xE _{0}$

  3. $x^{2}E _{0}$

  4. -$x^{2}E _{0}$

Show answer
Correct Option: B

The potential $V$ is varying with x and y as $\displaystyle V = \dfrac{1}{2}(y^2-4x)$ volt. The field at $x = 1 m , y = 1 m$, is :

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $\displaystyle 2\hat{i}+\hat{j} \ Vm^{-1}$

  2. $\displaystyle -2\hat{i}+\hat{j} \ Vm^{-1}$

  3. $\displaystyle 2\hat{i}-\hat{j} \ Vm^{-1}$

  4. $\displaystyle -2\hat{i}+2\hat{j} \ Vm^{-1}$

Show answer
Correct Option: C
Explanation:

$\displaystyle E _x = -\dfrac{dV}{dx} = -\dfrac{1}{2} [-4] = 2$

 $\displaystyle E _y = -\dfrac{dV}{dy} = -\dfrac{1}{2}[2y] = - y = -1$

 $\displaystyle \therefore \vec {E} = E _x \hat{i}+E _y \hat{j}=2\hat{i}-1\hat{j}$

The electric potential $V$ at any point $(x,y,z)$ in space is given by $V=4x^2$ volt. The electric field at $(1,0,2)$m in $Vm^{-1}$ is

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $8$, along the positive x-axis

  2. $8$, along the negative x-axis

  3. $16$, along the x-axis

  4. $16$, along the z-axis

Show answer
Correct Option: B

An electric field is expressed as $\displaystyle \vec{E} = 2\hat{i} + 3 \hat{j}$. Find the potential difference $(V _A - V _B)$ between two points $A$ and $B$ whose position vectors are given by $\displaystyle \vec r _A = \hat{i} + 2\hat{j}$ and $\displaystyle \vec r _B = 2\hat{i} + \hat{j}+3\hat{k}$ :

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $-1 V$

  2. $1 V$

  3. $2 V$

  4. $3 V$

Show answer
Correct Option: A
Explanation:

$dV=-\vec E.\vec dx-\vec E.\vec dy$
$\Delta V=-\int Edx-\int Edy$
$\displaystyle V _B-V _A = -(\int _{1}^{2}2dx+\int _{2}^{1}3dy)$

$\displaystyle =-[2(2-1)+3(1-2)]$
$\displaystyle =-[2-3] = 1 V$
 Hence, $V _A-V _B = -1 V$

An infinite nonconducting sheet of charge has a surface charge density of $10^{-7}\ C/m^2$. The separation between two equipotential surfaces near the sheet whose potential differ by $5\ V$ is

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $0.88\ cm$

  2. $0.88\ mm$

  3. $0.88\ m$

  4. $5\times 10^{-7}\ m$

Show answer
Correct Option: B

The electric potential V is given as a function of distance by $V=(5x^2+10x-4)volt$, where x is in metre. Value of electric field at $x=1m$ is :

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $-23 V/m$

  2. $11 V/m$

  3. $6 V/m$

  4. $-20 V/m$

Show answer
Correct Option: D
Explanation:

Given, $V=5x^2+10x-4$

or $\dfrac{dV}{dx}=10x+10$
The field, $E=-\dfrac{dV}{dx}=-(10x+10)$

At $x=1,  E=-(10+10)=-20 V/m$
So option D is correct. 

The potential at a point x (measured in $\mu m$) due to some charges situated on the x-axis is given by $V(x)=20/(x^2-4)volt$
The electric field E at $x=4\mu m$ is given by :

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $(10/9)volt /\mu m$ and in the $+ve$ x direction

  2. $(5/3)volt /\mu m$ and in the $-ve$ x direction

  3. $(5/3)volt /\mu m$ and in the $+ve$ x direction

  4. $(10/9)volt /\mu m$ and in the $-ve$ x direction

Show answer
Correct Option: A
Explanation:

Given, $\displaystyle V(x)=\dfrac{20}{x^2-4}$

Electric field , $\displaystyle E=-\dfrac{dV}{dx}=-\dfrac{-20}{(x^2-4)^2}(2x)=\dfrac{40x}{(x^2-4)^2}$

At $\displaystyle x= 4 \mu m,   E=\dfrac{40(4)}{(4^2-4)^2}=\dfrac{160}{144}=(10/9)  volt/\mu m$

Positive sign indicates that $\vec{E}$ is in the +ve x-direction.

A and B are two points in an electric field. If the work done in carrying $4.0 C$ of electric charge from A to B is $16.0 J$, the potential difference between A and B is :

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $zero$

  2. $2.0 V$

  3. $4.0 V$

  4. $16.0 V$

Show answer
Correct Option: C
Explanation:

The work done, $W _{A\rightarrow B}=q\int _{V _A}^{V _B}dV=q(V _B-V _A)$
or $16=4(V _B-V _A) \Rightarrow V _B-V _A=4  V$