Tag: coulomb's law

Questions Related to coulomb's law

Determine the electric field strength vector if the potential of the field depends on x, y coordinates as $V = a (x^2 - y^2)$, where a is a constant.

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

  1. $\vec{E} = - 2a(x\widehat{i} - y\widehat{j})$

  2. $\vec{E} = - a(x\widehat{i} - y\widehat{j})$

  3. $\vec{E} = - \dfrac{a(x\widehat{i} - y\widehat{j})}{2}$

  4. $\vec{E} = - \dfrac{a(x\widehat{i} - y\widehat{j})}{4}$

Show answer
Correct Option: A
Explanation:

$ \vec E = - \triangledown V $   ( negative of gradient of V)

$\vec E = - (\dfrac{dV}{dx} \hat i + \dfrac{dV}{dy} \hat j )=-a(2x \hat i +2y \hat j )  $

Determine the electric field strength vector if the potential of the field depends on x, y coordinates as $V = axy$ , where $a$ is a constant.

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

  1. $\vec{E} = -a(y\widehat{i} + y\widehat{j})$

  2. $\vec{E} = -a(x\widehat{i} + y\widehat{j})$

  3. $\vec{E} = -a(x\widehat{i} + x\widehat{j})$

  4. $\vec{E} = -a(y\widehat{i} + x\widehat{j})$

Show answer
Correct Option: D
Explanation:

$ \vec E = - \triangledown V $   ( negative of gradient of V)

$\vec E = - (\dfrac{dV}{dx} \hat i + \dfrac{dV}{dy} \hat j )=-a(y \hat i + x \hat j ) $

The electric potential existing in space is $V(x, y, z) = A (xy+ yz + zx)$. Find the expression for 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. $-A{(x + Z) \widehat{i} + (y + Z) \widehat{j} + (x + y) \widehat{k}}$

  2. $-A{(y + Z) \widehat{i} + (x + Z) \widehat{j} + (x + y) \widehat{k}}$

  3. $-Ax{ \widehat{i} +y \widehat{j} + Z\widehat{k}}$

  4. $-A{(x+y) \widehat{i} + (x + y) \widehat{j} + (x + y-2Z) \widehat{k}}$

Show answer
Correct Option: B
Explanation:

$ \vec E = - \triangledown V = -A[(y+z) \hat i + ( z+x) \hat j +( y+x) \hat k ] V/m $

At a certain distance from a point charge, the field intensity is 500 V/m and the potential is 3000 V. The distance and the magnitude of the charge respectively are :

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

  1. 6 m and 6 $\mu $C

  2. 4 m and 2 $\mu$C

  3. 6 m and 4 $\mu$C

  4. 6 m and 2 $\mu$C

Show answer
Correct Option: D
Explanation:

The electric field at distance d due to point charge q is $E=kq/d^2 $ and potential $V=kq/d$ 

so, $E=V/d $ or $ d=\dfrac{V}{E}=\dfrac{3000}{500}=6 m$

since, $V=kq/d $

or $3000=9\times 10^9\times \dfrac{q}{6} $

or $q=2\times 10^{-6} C=2 \mu C$

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

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

Show answer
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 $r$.