Tag: relation between electric field and electric potential

Questions Related to relation between electric field and electric potential

In a certain region of space, the electric potential is $V (x, y, z) = Axy - Bx^2$ $+Cy$, where $A, B\ and\ C$ are positive constants. Calculate the $x, y\ and\ z$ components of the electric field.

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

  1. $E _x = - Ax, E _y = -Ay, E _z = 0$

  2. $E _x = - Ax + 2Bx, E _y = -Ay -C, E _z = 0$

  3. $E _x = - Ay + 2Bx, E _y = -Ax -C, E _z = 0$

  4. $E _x = - Ay, E _y = -Ax, E _z = 0$

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Correct Option: C
Explanation:

Here, $V(x,y,z)=Axy-Bx^2+Cy$


So, $E _x=-\dfrac{V}{dx}=-[Ay-2Bx]=2Bx-Ay$;


$E _y=-\dfrac{dV}{dy}=-[Ax+C]=-Ax-C$ and 

$E _z=-\dfrac{dV}{dz}=0$

Potential difference between centre and surface of the sphere of radius R and uniform volume charge density $\rho$ within it 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. $\displaystyle \dfrac{\rho R^2}{6 \varepsilon _0}$

  2. $\displaystyle \dfrac{\rho R^2}{4 \varepsilon _0}$

  3. $\displaystyle \dfrac{\rho R^2}{3 \varepsilon _0}$

  4. $\displaystyle \dfrac{\rho R^2}{2 \varepsilon _0}$

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Correct Option: A
Explanation:

Using Gauss's law the electric field inside the sphere , $E.4\pi r^2=\dfrac{q}{\epsilon _0}=\rho\dfrac{(4/3)\pi r^3}{\epsilon _0}$

or $E=\dfrac{\rho r}{3\epsilon _0}$

Potential difference between surface and center is $V=-\int _R^0 E.dr=-\int _R^0 \dfrac{\rho r}{3\epsilon _0} dr=\dfrac{\rho R^2}{6\epsilon _0}$

A uniform electric field exists in x-y plane. The potential of points A (-2m, 2m), B(+2m, 2m) and C(2m, 4m) are 4 V, 16V and 12 V respectively. The electric field is :

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

  1. $(4\widehat{i} + 5 \widehat{j}) V/m$

  2. $(3\widehat{i} + 4 \widehat{j}) V/m$

  3. $-(3\widehat{i} + 4 \widehat{j}) V/m$

  4. $(3\widehat{i} - 4 \widehat{j}) V/m$

Show answer
Correct Option: D
Explanation:
Let equation of potential be $ax+by+c$ where $(x,y)$ are the co-ordinates of the point in $x, y$ plane.
so, $a(2)+b(2)+c=4$
$a(-2)+b(2)+c=16$
$a(2)+b(4)+c=12$
Solving above equations, we get $a=-3;b=4;c=2$
so equation of potential is $V=-3x+4y+2$ 
Now the electric field is $\vec E=(-dV/dx)\vec i+(-dV/dy)\vec j=3\vec i-4\vec j$

In a certain region of space, the electric potential is $V (x, y, z) = Axy - Bx^2$ $+Cy$, where $A, B\ and\ C$ are positive constants. At which points is the electric field equal to zero?

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

  1. $x = +C/A, y = +BC/A$ $^2$, any value of $z$

  2. $x = +C/A, y = +2BC/A$ $^2$, any value of $z$

  3. $x = -C/A, y = -2BC/A$ $^2$, $z=0$

  4. $x = -C/A, y = -2BC/A$ $^2$, any value of $z$

Show answer
Correct Option: D
Explanation:

$ \vec E = - \triangledown V = - [ ( Ay - 2Bx) \hat i + ( Ax + C) \hat j ] $


$ \vec E = 0$  at, 

$ E _y = 0 \Rightarrow Ax + C= 0 \Rightarrow x = -C/A $

$ E _x= 0 \Rightarrow Ay - 2Bx= Ay + 2BC/A =0 \Rightarrow y = -2BC/A^2 $

$ E _z = 0 $  everywhere . 

The electric potential existing in space is $V(x, y, z) = A (xy+ yz + zx)$. If A is $10$ SI units, find the magnitude of the electric field at $(1 m, 1 m, 1 m)$ :

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

  1. $20 \sqrt 2$ N/C

  2. $20 \sqrt 3$ N/C

  3. $10 \sqrt 3$ N/C

  4. $20 $ N/C

Show answer
Correct Option: B
Explanation:

$ \vec E = - \triangledown V = -A[(y+z) \hat i + ( z+x) \hat j +( y+x) \hat k ] = -10[2 \hat i + 2 \hat j + 2 \hat k ] = 20\sqrt{3} N/C$

The electric potential at a point (x, y) in the x-y plane is given by V = - Kxy. The field intensity at a distance r in this plane, from the origin is proportional to :

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

  1. $r^2$

  2. $r$

  3. $1/r$

  4. $1/r^2$

Show answer
Correct Option: B
Explanation:
Let the $x-y$ coordinates of the point at distance $r$ from the origin be given as $x=rcos\theta$, $y=rsin\theta$
Potential is given as $V=-Kxy$
Now, Electric field intensity is $\vec E=(-dV/dx)\vec i+(-dV/dy)\vec j=K(y\vec i+x\vec j)=K(rsin\theta\vec i+rcos\theta\vec j)=Kr(sin\theta\vec i+cos\theta\vec j)\propto r$
So electric field potential is proportional to $r$.

Find out the relationship between the electric field and electric potential include which of the following statement?

I. If the electric field at a certain point is zero, then the electric potential at the same point is also zero.

II. The electric potential is inversely proportional to the strength of the electric field.

III. If the electric potential at a certain point is zero, then the electric field at the same point is also zero.

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

  1. I only

  2. II only

  3. I and II only

  4. I and III only

  5. None of the above

Show answer
Correct Option: E
Explanation:

If electric field at some point is zero, it is not necessary for the electric potential to be the same. Consider the mid point of line joining to equal charges of same sign. Field there is zero, but potential is finite and positive.


Electric field strength is given by $E=\dfrac{kQq}{r^2}$
Electric potential is given by $V=\dfrac{dQq}{r}$
Clearly they are not inversely proportional to each other.

If electric potential at some point is zero, it is not necessary for the electric field to be the same. Consider the mid point of line joining to equal charges of opposite signs. Potential there is zero, but electric field exists.

When negative charges are kept in electric field then negative charges are accelerated by electric fields toward points:

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

  1. at lower electric potential

  2. at higher electric potential

  3. where the electric field is zero

  4. where the electric field is weaker

  5. where the electric field is stronger

Show answer
Correct Option: B
Explanation:

We know that , $F=-\Delta U$, change in potential energy. 

As charge is negative so, $F=-qE=-\Delta U$ or $\Delta U=qE$
So we will get change in potential as positive and hence the electric field should be towards  points at higher potential.

An electric field (in $V/m$) is given by $E=10x^3$. Determine the potential difference, in volts, between $x=0m$ and $x=3m$.

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

  1. $202.5$

  2. $100$

  3. $20$

  4. $250$

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Correct Option: A
Explanation:

We know that, $E=-\dfrac{dV}{dx}$


$V=-\int _0^3 Edx=-\int _0^3 10x^3 dx=-10\times \dfrac{3^4}{4}=202.5 $

In the direction of electric field, the electric potential:

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

  1. Decreases

  2. Increases

  3. Remains unchanged

  4. Becomes zero

Show answer
Correct Option: A
Explanation:

$dV=-E.dr$, so the electric potential $V$ decreases continuously as we move along the direction of the electric field