Tag: coulomb's law

Questions Related to coulomb's law

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

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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

The potential at a point $x$ (measured in $\mu m )$ due to somecharges situated on the $x$ -axis is given by $V ( x ) = 20 / \left( x ^ { 2 } - 4 \right)$Volts. 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. 5$/ 3$ Volt / \mum and in the -ve $x$ direction

  2. 5$/ 3$ Volt $/ \mu m$ and in the +ve $x$ direction

  3. 10$/ 9$ Volt / \mum and in the -ve $x$ direction

  4. 10$/ 9$ Volt $/ \mu m$ and in the +ve $x$ direction

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Correct Option: B

Variation in potential is maximum if one goes :

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

  1. along the line of force

  2. perpendicular to the line of force

  3. in any direction

  4. none of these

Show answer
Correct Option: A
Explanation:

$dV=-\vec{E}.d\vec{r}=-Edr cos\theta$


Hence, variation will be maximimum for $\theta=0^{o}$ or $180^{o}$, that is variation $dV$ is maximum along line of field or say line of force.

Answer-(A)

The electric field lines are closer together near object $A$ than they are near object $B$. We can conclude that :

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 near $A$ is greater than the potential near $B$

  2. the potential near $A$ is less than the potential near $B$

  3. the potential near $A$ is equal to the potential near $B$

  4. nothing about the relative potentials near $A$ and $B$

Show answer
Correct Option: D
Explanation:

Potential decreases in the direction of electric field. So it depends  on whether the lines of forces are from $A$ to $B$ or from $B$ to $A$.

There is an electric field $E$ in the x-direction. If the work done by the electric field in moving a charge of $0.2 C$ through a distance of $2 m$ along a line making an angle $60^{\circ}$ with the x-axis is $4 J$, then what is the value of $E$?

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

  1. $\displaystyle \sqrt3 NC^{-1}$

  2. $\displaystyle 4 NC^{-1}$

  3. $\displaystyle 5 NC^{-1}$

  4. $\displaystyle 20 NC^{-1}$

Show answer
Correct Option: D
Explanation:

$\displaystyle F = qE$
work will only be done in moving the charged particle in $x$ direction

work done in moving the charge in y-direction will be $0$
Work done , $W=\int \vec{F}.\vec{dr}$

$\displaystyle W = qE \times 2 cos  60^{\circ}$

or $\displaystyle 4 = 0.2E\times 2 \times \dfrac{1}{2}$

$  \implies  E = 20 NC^{-1}$

Charge $Q$ is given a displacement $\displaystyle \vec{r} = a\hat{i}+b\hat{j}$ in an electric field $\displaystyle \vec{E} = E _1\hat{i}+E _2\hat{j}$. The work done is :

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

  1. $\displaystyle Q(E _1a+E _2b)$

  2. $\displaystyle Q\sqrt{(E _1a)^2+(E _2b)^2}$

  3. $\displaystyle Q (E _1+E _2) \sqrt{a^2+b^2}$

  4. $\displaystyle Q \sqrt{(E _1^2+E^2 _2)^2} \sqrt{a^2+b^2}$

Show answer
Correct Option: A
Explanation:

Work done in the presence of electric field E is $W=\vec F. \vec r = q\vec E.\vec r$
$W=Q[(E _1\hat{i}+E _2\hat{j}).( a\hat{i}+b\hat{j})]$
$W=Q(E _1a+E _2b)$

The electric potential decreases uniformly from $120V$ to $80V$ as one moves on the x-axis from $x=-1cm$ to $x=+1cm$. 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 $20V{cm}^{-1}$

  2. may be equal to $20V{cm}^{-1}$

  3. may be greater than $20V{cm}^{-1}$

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

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Correct Option: B,C