Q: Equipotential surfaces are shown in fig, then the electric field strength will be

Using\$,\$ \$dV = - \overrightarrow E \cdot \overrightarrow {dr} \$\$\Delta V = - E \cdot \Delta rcs\theta \$\$E = \frac{{ - \Delta V}}{{\Delta r\cos \theta }}\$\$E = \frac{{ - \left( { - 20 - 10} \right)}}{{10 \times {{10}^{ - 2}}\cos {{120}^0}}} = \frac{{ - 10}}{{10 \times {{10}^{ - 2}}\left( { - \sin {{30}^0}} \right)}} = \frac{{ - {{10}^2}}}{{ - 1/2}} = 200\,V/m\$direction of \$E\$ be perpendicular to the equipotential surface \$i.e.,\$ at \${120^0}\$ with \$x-\$ axis\$.\$Hence,option \$(C)\$ is correct answer.

Q: Two point charges +9q and +q are kept 16 cm apart. Where should a third charge Q be placed between them so that the system remains in equilibrium ?

Two point charges \$+9q\$ and \$+9\$. Kept apart \$=16cm=16\times { 10 }^{ -2 }m\$Two cases will be needed.Given charges, \$+q\$, \$+9q\$. let, the third charge \$q\$ be placed at a distance \$x\$ from \$q\$ and hence its distance from \$9q\$ and \$(16-x)cm\$ for the system to remain in equilibrium \${ F }_{ 1 }={ F }_{ 2 }\$Therefore, \${ F }_{ 1 }=\dfrac { K\left( q\times Q \right) }{ { x }^{ 2 } } \$\${ F }_{ 2 }=\dfrac { K\left( 9q\times Q \right) }{ { \left( 16-x \right) }^{ 2 } } \$Equating \${ F }_{ 1 }\$ and \${ F }_{ 2 }\$ we get,\$\dfrac { 9 }{ { x }^{ 2 } } =\dfrac { 1 }{ { \left( 16-x \right) }^{ 2 } } \$\${ F }_{ 1 }={ F }_{ 2 }\$or, \$\dfrac { K\left( q\times Q \right) }{ { x }^{ 2 } } =\dfrac { K\left( 9q\times Q \right) }{ { \left( 16-x \right) }^{ 2 } } \$or, \$\dfrac { 1 }{ { x }^{ 2 } } =\dfrac { 9 }{ { \left( 16-x \right) }^{ 2 } } \$or, \$\dfrac { 1 }{ x } =\dfrac { 9 }{ \left( 16-x \right) } \$or, \$16-x=3x\$or, \$4x=16\$or, \$x=4\$\$\therefore\$ \$(16-4)cm=12cm\$ from \$+9q\$.

Q: In a region of constant potential:

In a region of constant potential\$E=-\cfrac { dV }{ dr } =0\$ (\$\because\$ \$V=\$ constant)i.e., electric field is zero, so there can be no charge inside the region.

Q: A uniformly charged solid sphere of radius R has potential \$V_0\$ (measured with respect to \$\infty\$) on its surface. For this sphere the equipotential surfaces with potentials \$\displaystyle \frac{3V_0}{2}, \frac{5 V_0}{4}, \frac{3V_0}{4}\$ and \$\displaystyle \frac{V_0}{4}\$ have radius \$R_1, R_2, R_3\$ and \$R_4\$ respectively. Then :

\$V(R) = \dfrac{kQ}{R} = V_0\$We use results derived for potential.\$V(r)= \dfrac{kQ}{r}\$ for \$r \gt R\$\$V(r) = \dfrac{kQ}{2R^3}(3R^2-r^2)\$\$V(r = 0)= \dfrac{3kQ}{2R} = \dfrac{3V_0}{2}\$Hence, \$V(R_1)\$ matches with V at center. \$ \therefore R_1=0\$Thus, we eliminate option(B).At \$R_2\$\$V(R_2) =\dfrac{5V_0}{4} = \dfrac{kQ}{2R^3}(3R^2-R_2^2)\$\$ \Rightarrow \dfrac{5V_0}{4} = \dfrac{1}{2} ( \dfrac{kQ}{R})( 3-\dfrac{R_2^2}{R^2})= \dfrac{1}{2} V_0 ( 3-\dfrac{R_2^2}{R^2})\$\$ \Rightarrow \dfrac{5}{2}= 3-\dfrac{R_2^2}{R^2}\$\$ \Rightarrow \dfrac{1}{2} = \dfrac{R_2^2}{R^2}\$\$ \Rightarrow R_2 = \dfrac{R}{\sqrt{2}}\$For \$R_3\$\$V(R_3) = \dfrac{3V_0}{4} = \dfrac{kQ}{R_3}\$\$ \Rightarrow \dfrac{3V_0}{4} = V_0 \dfrac{R}{R_3}\$\$ \Rightarrow R_3 = \dfrac{4R}{3}\$For \$R_4\$\$V(R_4) = \dfrac{V_0}{4} = \dfrac{kQ}{R_4}\$\$ \Rightarrow R_4 =4R\$Thus, \$R_4 -R_3 = 4R - \dfrac{4R}{3}= \dfrac{8R}{3}\$\$R_2 = \dfrac{R}{\sqrt{2}} \lt \dfrac{8R}{3}=R_4 -R_3\$Also, \$2R \lt 4R = R_4\$

Q: Two small spheres each carrying a charge q are placed r metre apart. If one of the spheres is taken around the other one in a circular path, the work done will be equal to :

Potential energy between any two given particles\$=\cfrac{kq_1q_2}{r}\$since revolving in circle does not change the distance \$r\$ between these charges, change in P.E. will be equal to zerohence \$work\$ \$done=0\$

Question 1

Equipotential surfaces are shown in fig, then the electric field strength will be

Accepted Answer

Using$,$ $dV =  - \overrightarrow E  \cdot \overrightarrow {dr} $$\Delta V =  - E \cdot \Delta rcs\theta $$E = \frac{{ - \Delta V}}{{\Delta r\cos \theta }}$$E = \frac{{ - \left( { - 20 - 10} \right)}}{{10 \times {{10}^{ - 2}}\cos {{120}^0}}} = \frac{{ - 10}}{{10 \times {{10}^{ - 2}}\left( { - \sin {{30}^0}} \right)}} = \frac{{ - {{10}^2}}}{{ - 1/2}} = 200\,V/m$direction of $E$ be perpendicular to the equipotential surface $i.e.,$ at ${120^0}$ with $x-$ axis$.$Hence,option $(C)$ is correct answer.

Question 2

Two charges $2\mu C$ and $-2\mu C$ are placed at points A and B 6 cm apart.(a) Identify an equipotential surface of the system.(b) What is the direction of the electric field at every point on this surface?

Accepted Answer

(a) An equipotential surface is the plane on which total potential is zero.It is at the midpoint of the line AB and perpendicular to it.(b) The direction of the electric field at every point on this surface is normal to the plane in direction of AB.

Question 3

Define equipotential surface.

Accepted Answer

Any surface over which the potential is constant is called an equipotential surface.In other words, the potential difference between any two points on an equipotential surface is zero.Some important properties of equipotential surfaces :1. Work done in moving a charge over an equipotential surface is zero.2. The electric field is always perpendicular to an equipotential surface.3. The spacing between equipotential surfaces enables us to identify regions of strong and weak fields.4. Two equipotential surfaces can never intersect. If two equipotential surfaces could intersect, then at the point of intersection there would be two values of electric potential which is not possible.

Question 4

Two point charges +9q and +q are kept 16 cm apart. Where should a third charge Q be placed between them so that the system remains in equilibrium ?

Accepted Answer

Two point charges $+9q$ and $+9$. Kept apart $=16cm=16\times { 10 }^{ -2 }m$Two cases will be needed.Given charges, $+q$, $+9q$. let, the third charge $q$ be placed at a distance $x$ from $q$ and hence its distance from $9q$ and $(16-x)cm$ for the system to remain in equilibrium ${ F }_{ 1 }={ F }_{ 2 }$Therefore, ${ F }_{ 1 }=\dfrac { K\left( q\times Q \right)  }{ { x }^{ 2 } } $${ F }_{ 2 }=\dfrac { K\left( 9q\times Q \right)  }{ { \left( 16-x \right)  }^{ 2 } } $Equating ${ F }_{ 1 }$ and ${ F }_{ 2 }$ we get,$\dfrac { 9 }{ { x }^{ 2 } } =\dfrac { 1 }{ { \left( 16-x \right)  }^{ 2 } } $${ F }_{ 1 }={ F }_{ 2 }$or,  $\dfrac { K\left( q\times Q \right)  }{ { x }^{ 2 } } =\dfrac { K\left( 9q\times Q \right)  }{ { \left( 16-x \right)  }^{ 2 } } $or,  $\dfrac { 1 }{ { x }^{ 2 } } =\dfrac { 9 }{ { \left( 16-x \right)  }^{ 2 } } $or,  $\dfrac { 1 }{ x } =\dfrac { 9 }{ \left( 16-x \right)  } $or,  $16-x=3x$or,  $4x=16$or,  $x=4$$\therefore$   $(16-4)cm=12cm$ from $+9q$.

Question 5

In a region of constant potential:

Accepted Answer

In a region of constant potential$E=-\cfrac { dV }{ dr } =0$ ($\because$ $V=$ constant)i.e., electric field is zero, so there can be no charge inside the region.

Question 6

Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximately:

Accepted Answer

A collection of charge located at a very large distance can b considered as the point charge. Now the equipotential surface for a point charge will have the same distance from the point in all the directions.Therefore, the euipotential points for a point charge will have a spherical surface.So, Option $A$ is correct.

Question 7

A uniformly charged solid sphere of radius R has potential $V_0$ (measured with respect to $\infty$) on its surface. For this sphere the equipotential surfaces with potentials $\displaystyle \frac{3V_0}{2}, \frac{5 V_0}{4}, \frac{3V_0}{4}$ and $\displaystyle \frac{V_0}{4}$ have radius $R_1, R_2, R_3$ and $R_4$ respectively. Then :

Accepted Answer

$V(R) = \dfrac{kQ}{R} = V_0$We use results derived for potential.$V(r)= \dfrac{kQ}{r}$ for $r \gt R$$V(r) = \dfrac{kQ}{2R^3}(3R^2-r^2)$$V(r = 0)= \dfrac{3kQ}{2R} = \dfrac{3V_0}{2}$Hence, $V(R_1)$ matches with  V at center. $ \therefore R_1=0$Thus, we eliminate option(B).At $R_2$$V(R_2) =\dfrac{5V_0}{4} = \dfrac{kQ}{2R^3}(3R^2-R_2^2)$$ \Rightarrow \dfrac{5V_0}{4} = \dfrac{1}{2} ( \dfrac{kQ}{R})( 3-\dfrac{R_2^2}{R^2})= \dfrac{1}{2} V_0 ( 3-\dfrac{R_2^2}{R^2})$$ \Rightarrow \dfrac{5}{2}= 3-\dfrac{R_2^2}{R^2}$$ \Rightarrow  \dfrac{1}{2} = \dfrac{R_2^2}{R^2}$$ \Rightarrow R_2 = \dfrac{R}{\sqrt{2}}$For $R_3$$V(R_3) = \dfrac{3V_0}{4} = \dfrac{kQ}{R_3}$$ \Rightarrow \dfrac{3V_0}{4}  = V_0 \dfrac{R}{R_3}$$ \Rightarrow R_3 = \dfrac{4R}{3}$For $R_4$$V(R_4) = \dfrac{V_0}{4} = \dfrac{kQ}{R_4}$$ \Rightarrow R_4 =4R$Thus, $R_4 -R_3 = 4R - \dfrac{4R}{3}= \dfrac{8R}{3}$$R_2 = \dfrac{R}{\sqrt{2}} \lt \dfrac{8R}{3}=R_4 -R_3$Also, $2R \lt 4R = R_4$

Question 8

Two small spheres each carrying a charge q are placed r metre apart. If one of the spheres is taken around the other one in a circular path, the work done will be equal to :

Accepted Answer

Potential energy between any two given particles$=\cfrac{kq_1q_2}{r}$since revolving in circle does not change the distance $r$ between these charges, change in P.E. will be equal to zerohence $work$ $done=0$

Question 9

Angle between an equipotential surface and electric lines of force is :

Accepted Answer

The angle between Electric field and an equi-potential surface is always $90^0$. This is because,when the potential becomes constant,the negative potential gradient also becomes zero,hence necessitating the need for Electric field to be always normal with surface.

Question 10

What is the equipotential surface? Draw the equipotential surface due to point charge.

Accepted Answer

Equipotential SurfaceThe surface in an electric field where the value of electric potential is the same at all the points on the surface is called equipotential surface. The potential difference between two points in an equipotential surface is zero. Thus, work done in moving the charge from one point to another in an equipotential surface is zero. since the work is done is zero when the direction of force (electric field) and displacement is normal. Thus, electric field is perpendicular to the plane in an equipotential surface. It is to be noted for the equipotential surface:1. For a uniform electric field $ \vec{E} $, the equipotential surface is normal to its field lines. According to figure 3.8,1(a) II and III are equipotential surfaces.2. For an isolated point charge:Potential at a distance r due to point charge $ +\mathrm{q} $$V=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{q}{r}$If we draw a sphere of radius r surrounding the $ + $ q charge. Then, potential will be same at all points on the sphere i.e., equipotential surfaces are spherical whose charge is at the centre. For a positive point charge, as the radius of the spherical surface increases, the potential of spherical surface decreases.3. Equipotential surface for an electric dipole: The equipotential surface for an electric dipole are shown dotted in figure 3.8 ( $ \mathrm{c} $ ).4. Equipotential surfaces for a pair of similar point charges: The equipotential surface is as shown in figure$ 3.8(\mathrm{d}) $

Equipotential Surface of Various Charge System - I

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Equipotential Lines And Equipotential Surfaces

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Equipotential surface and its properties