Tag: field strength and potential gradient

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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
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 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.

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

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

Mark the correct statement:

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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).

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

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 .......

  1. $xE _{0}$
  2. -$xE _{0}$
  3. $x^{2}E _{0}$
  4. -$x^{2}E _{0}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a uniform field E = E0, V = -integral(E dx) = -E0 * x + C. Since V(0) = 0, C = 0. Thus, V(x) = -E0 * x.

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 $V$ 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 :

  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. $\displaystyle -2\hat{i}+2\hat{j} \ Vm^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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