Tag: coulomb's law

Quizzes (4) Questions (102)

Questions Related to coulomb's law

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 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 :

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $F=qE _x=1(8)=8 N$

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 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 :

  1. $zero$
  2. $qEB$
  3. $qEa$
  4. $-qEa$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 , $W=\int \vec{F}.\vec{dr}=\int qE\hat i.(\hat{i}dx+\hat{j}dy+\hat{k}dz)$
$W=qE\int dx=qE[\int _a^{2a}dx+\int^a _{2a}dx+\int _a^{0}dx]=qE[2a-a+a-2a+0-a]=-qEa$

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

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 = 

  1. $k\sqrt{6}$
  2. $2k\sqrt{6}$
  3. $2k\sqrt{3}$
  4. $4k\sqrt{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The electric field is E = -grad(V). E = -(dV/dx i + dV/dy j + dV/dz k). Given V = k(2x^2 - y^2 + z^2), E = -k(4x i - 2y j + 2z k). At (1, 1, 1), E = -k(4 i - 2 j + 2 k). The magnitude is k * sqrt(4^2 + (-2)^2 + 2^2) = k * sqrt(16 + 4 + 4) = k * sqrt(24) = 2k * sqrt(6).

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 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

  1. $10 V$
  2. $90 V$
  3. $1000 V$
  4. $9000 V$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The electric field E = F/q = 3000 N / 3 C = 1000 N/C. The potential difference V = E * d. With d = 1 cm = 0.01 m, V = 1000 * 0.01 = 10 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

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

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

The electric field in the x-direction is given by E = -dV/dx. Given V(x) = 20 / (x^2 - 4), differentiating with respect to x using the quotient rule gives dV/dx = 20 * (-1) * (2x) / (x^2 - 4)^2 = -40x / (x^2 - 4)^2. At x = 4, dV/dx = -40(4) / (16 - 4)^2 = -160 / (12^2) = -160 / 144 = -10/9. Since E = -dV/dx, E = -(-10/9) = +10/9 Volt/um, and the positive sign indicates it is in the positive 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

Variation in potential is maximum if one goes :

  1. along the line of force

  2. perpendicular to the line of force

  3. in any direction

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
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)

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

  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 $B$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $A$.

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

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 $E$?

  1. $\displaystyle \sqrt3 NC^{-1}$
  2. $\displaystyle 4 NC^{-1}$
  3. $\displaystyle 5 NC^{-1}$
  4. $\displaystyle 20 NC^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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}$

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

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 :

  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. $\displaystyle Q \sqrt{(E _1^2+E^2 _2)^2} \sqrt{a^2+b^2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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 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

  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 $20V{cm}^{-1}$
Reveal answer Fill a bubble to check yourself
B,C Correct answer