Tag: relation between electric field and electric potential

Questions Related to relation between electric field and electric potential

The electric field in a region is directed outward and is proportional to the distance r from the origin. Taking the electric potential at the origin to be zero,

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

  1. it is uniform in the region

  2. it is proportional to r

  3. it is proportional to r$^2$

  4. it increases as one goes away from the origion

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Correct Option: C
Explanation:

Electric field is directly propotional to $r$, therefore
$E=kr$
We know that
$\ V=-\int _{ 0 }^{ r }{ \overrightarrow { E. } \overrightarrow { dr }  } $
which gives 
$V=-\dfrac { k{ r }^{ 2 } }{ 2 } $ since V at $r=0$ is $0$
Hence V is proportional to ${ r }^{ 2 }$

A uniform electric field of $20$ NC$^{-1}$ exists along the x-axis in space. The potential difference V$ _B-$V$ _A$ for the point A $=$ $(4 m, 2m)$ and B $=$ $(6m, 5m)$ is:

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

  1. $20$ $\sqrt{13}$ V

  2. $- 40 V$

  3. zero V

  4. none of the above

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Correct Option: B
Explanation:

Here, $\vec{E}=20 \hat i \Rightarrow E _x=20$
Potential difference ,$V _B-V _A=-\int _A^B Edr=-\int _4^6E _x dx=-20\int _4^6 dx=-20(6-4)=-40  V$

The electric field at the origin is along the positive X-axis. A small circle is drawn with the centre at the origin cutting the axes at points A, B, C and D having coordinates $(a, 0), (0, a), (-a, 0), (0, -a)$ respectively. Out of the given points on the periphery of the circle, the potential is minimum at :

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

  1. A

  2. B

  3. C

  4. D

Show answer
Correct Option: A
Explanation:

The relation between electric field and potential is given by $\vec{E} = \displaystyle -\frac{\partial V}{\partial x}\hat{i} -\frac{\partial V}{\partial y}\hat{j} $

Given that, at origin, electric field is along positive x-axis.
Thus, $\displaystyle \frac{\partial V}{\partial x} < 0$ and $\displaystyle \frac{\partial V}{\partial y} = 0$
Thus, $V$ decreases in the positive x-direction and remains constant in y-direction.
Hence, minimum $V$ occurs at $(a,0)$ i.e., $A$

It is found that air breaks down electrically, when the electric field is $  3 \times 10^{6} \mathrm{V} / \mathrm{m} .  $ What is the potential to which a sphere of radius $1  \mathrm{m}  $ can be raised, before sparking takes place?

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

  1. $ V=10^{6} \mathrm{V} $

  2. $ V=2 \times 10^{6} \mathrm{V} $

  3. $ V=3 \times 10^{6} \mathrm{V} $

  4. $ V=4 \times 10^{6} \mathrm{V} $

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Correct Option: C

In moving from A to B along an electric field line, the wok done by the electric field on an electron is $6.4 \times 10^{-19}$ J. If $\phi _1$ and $\phi _2$ are equipotential surfaces, then the potential difference $V _b-V _A $ is

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

  1. -4V

  2. 4V

  3. zero

  4. 6.4 V

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Correct Option: B

The equation of an equipotential line is an electric field is y = 2x, then the electric field strength vector at (1, 2) may be 

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

  1. $4\vec{i} + 3\vec{j}$

  2. $4\vec{i} + 8\vec{j}$

  3. $8\vec{i} + 4\vec{j}$

  4. $-8\vec{i} + 4\vec{j}$

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Correct Option: B

The electric potential in a certain region along the x-axis varies with x according to the relation $V(x) = 5 - 4x^2$. Then, the correct statement is :

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

  1. the potential difference between the points $x =1$m and $x=2$m is $12$ Volt

  2. the force experienced by a Coulomb of charge placed at $x =1$ m is $8$ Newton

  3. the electric field components along Y and Z direction are zero

  4. all of the above

Show answer
Correct Option: D
Explanation:

$V(x)=5-4x^2$

$V(1)=5-4=1 V,  V(2)=5-4(2^2)=-11 V$

Potential difference between $x=1 m$ and $x=2 m$ is $V _{12}=V _1-V _2=1-(-11)=12 V$

here, $E _x=-\dfrac{dV}{dx}=8x,  E _y=-\dfrac{dV}{dy}=0$ and $E _z=-\dfrac{dV}{dz}=0$

The electric force on $1$ coulomb charge at $x=1$ is $F=qE _x=1(8)=8 N$

A point charge q moves from point P to a point S along a path PQRS in a uniform electric field E pointing parallel to the x-axis. The coordinates of P, Q. R and S are $(a, b, 0), (2a, 0, 0), (a, -b, 0)$ and $(0, 0, 0)$. The work done by the field in the above process 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. $qEB$

  3. $qEa$

  4. $-qEa$

Show answer
Correct Option: D
Explanation:

As the field E is uniform, so E is constant at every point.
As E is directed parallel to x axis, so $\vec{E}=E\hat i$
The work done , $W=\int \vec{F}.\vec{dr}=\int qE\hat i.(\hat{i}dx+\hat{j}dy+\hat{k}dz)$
$W=qE\int dx=qE[\int _a^{2a}dx+\int^a _{2a}dx+\int _a^{0}dx]=qE[2a-a+a-2a+0-a]=-qEa$

In a certain region of space, the potential is given by : $V = k {[2x^2 - y^2 + z^2]}$. The electric field at the point (1, 1, 1) has magnitude = 

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

  1. $k\sqrt{6}$

  2. $2k\sqrt{6}$

  3. $2k\sqrt{3}$

  4. $4k\sqrt{3}$

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Correct Option: B

A charge of 3C moving in a uniform electric field experiences a force of $3000 N$. The potential difference between two points situated in the field at a distance $1 cm$ from each other 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. $10 V$

  2. $90 V$

  3. $1000 V$

  4. $9000 V$

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Correct Option: A