Tag: relation between electric field and electric potential

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Questions Related to relation between electric field and electric potential

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

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

  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
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

E = -dV/dx. V = 4x^2, so E = -d(4x^2)/dx = -8x. At x = 1, E = -8 V/m. The negative sign indicates the direction is along the negative x-axis.

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

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 $\displaystyle \vec r _B = 2\hat{i} + \hat{j}+3\hat{k}$ :

  1. $-1 V$
  2. $1 V$
  3. $2 V$
  4. $3 V$
Reveal answer Fill a bubble to check yourself
A Correct answer
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, $V _A-V _B = -1 V$

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

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

  1. $0.88\ cm$
  2. $0.88\ mm$
  3. $0.88\ m$
  4. $5\times 10^{-7}\ m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The electric field of an infinite sheet is E = sigma / (2 * epsilon_0). Given sigma = 10^-7 C/m^2 and epsilon_0 = 8.85 * 10^-12 F/m, E = 10^-7 / (2 * 8.85 * 10^-12) = 10^5 / 17.7 = 5649 V/m. Using V = E * d, d = V / E = 5 / 5649 = 0.000885 m = 0.88 mm.

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

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 :

  1. $-23 V/m$
  2. $11 V/m$
  3. $6 V/m$
  4. $-20 V/m$
Reveal answer Fill a bubble to check yourself
D Correct answer
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. 

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

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 :

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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 :

  1. $zero$
  2. $2.0 V$
  3. $4.0 V$
  4. $16.0 V$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

Determine the electric field strength vector if the potential of the field depends on x, y coordinates as $V = a (x^2 - y^2)$, where a is a constant.

  1. $\vec{E} = - 2a(x\widehat{i} - y\widehat{j})$
  2. $\vec{E} = - a(x\widehat{i} - y\widehat{j})$
  3. $\vec{E} = - \dfrac{a(x\widehat{i} - y\widehat{j})}{2}$
  4. $\vec{E} = - \dfrac{a(x\widehat{i} - y\widehat{j})}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ \vec E = - \triangledown V $   ( negative of gradient of V)

$\vec E = - (\dfrac{dV}{dx} \hat i + \dfrac{dV}{dy} \hat j )=-a(2x \hat i +2y \hat j )  $

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

Determine the electric field strength vector if the potential of the field depends on x, y coordinates as $V = axy$ , where $a$ is a constant.

  1. $\vec{E} = -a(y\widehat{i} + y\widehat{j})$
  2. $\vec{E} = -a(x\widehat{i} + y\widehat{j})$
  3. $\vec{E} = -a(x\widehat{i} + x\widehat{j})$
  4. $\vec{E} = -a(y\widehat{i} + x\widehat{j})$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$ \vec E = - \triangledown V $   ( negative of gradient of V)

$\vec E = - (\dfrac{dV}{dx} \hat i + \dfrac{dV}{dy} \hat j )=-a(y \hat i + x \hat j ) $

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

The electric potential existing in space is $V(x, y, z) = A (xy+ yz + zx)$. Find the expression for the electric field :

  1. $-A{(x + Z) \widehat{i} + (y + Z) \widehat{j} + (x + y) \widehat{k}}$
  2. $-A{(y + Z) \widehat{i} + (x + Z) \widehat{j} + (x + y) \widehat{k}}$
  3. $-Ax{ \widehat{i} +y \widehat{j} + Z\widehat{k}}$
  4. $-A{(x+y) \widehat{i} + (x + y) \widehat{j} + (x + y-2Z) \widehat{k}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ \vec E = - \triangledown V = -A[(y+z) \hat i + ( z+x) \hat j +( y+x) \hat k ] V/m $

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

At a certain distance from a point charge, the field intensity is 500 V/m and the potential is 3000 V. The distance and the magnitude of the charge respectively are :

  1. 6 m and 6 $\mu $C
  2. 4 m and 2 $\mu$C
  3. 6 m and 4 $\mu$C
  4. 6 m and 2 $\mu$C
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The electric field at distance d due to point charge q is $E=kq/d^2 $ and potential $V=kq/d$ 

so, $E=V/d $ or $ d=\dfrac{V}{E}=\dfrac{3000}{500}=6 m$

since, $V=kq/d $

or $3000=9\times 10^9\times \dfrac{q}{6} $

or $q=2\times 10^{-6} C=2 \mu C$