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: 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: 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 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\$

Q: Figure shows three spherical and equipotential surfaces \$A\$, \$B\$ and \$C\$ around a point charge \$q\$. The potential difference \${ V }_{ A }-{ V }_{ B }={ V }_{ B }-{ V }_{ C }\$. If \${ t }_{ 1 }\$ and \${ t }_{ 2 }\$ be the distances between them, then :

The potential around a point charge is \$V(r) = \displaystyle \frac{q}{4\pi\epsilon_0r}\$Let \$a,b,c\$ be the radii of \$A,B,C\$ respectively.Thus, \$t_1 = b-a\$ and \$t_2 = c-b\$\$V_A - V_B = \displaystyle \frac{q}{4\pi\epsilon_0a} - \frac{q}{4\pi\epsilon_0b} = \frac{q\ t_1}{4\pi\epsilon_0ab}\$\$V_B-V_C = \displaystyle \frac{q}{4\pi\epsilon_0b} - \frac{q}{4\pi\epsilon_0c} = \frac{q\ t_2}{4\pi\epsilon_0bc}\$But, \$V_A - V_B = V_B - V_C\$\$\Rightarrow \displaystyle \frac{V_A - V_B}{V_B - V_C} = 1\Rightarrow \frac{t_1}{ab}\times \frac{bc}{t_2}=1\Rightarrow\frac{t_1}{t_2} = \frac{a}{c}\$As \$a<c\$, we have \$t_1 < t_2\$

Q: If a unit charge is taken from one point to another over an equipotential surface, then work done on the charge is :

Equipotential surface has same potential at all points on the surface.Thus potential difference between two points is zero i.e. \$\Delta V = 0\$ Work done in moving a charge over an equipotential surface \$W = q\Delta V\$\$\implies\$ \$W = 0\$

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

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.

Q: Equipotential surfaces are shown in the figure. Find the magnitude and the direction of the resultant electric field.

Since, each of the equipotential surfaces are parallel to each other,the perpendicular distance between them is given byd \$= 10 sin 30 =\$ 5Now, the potential gradient between 2 consecutive planes is 10VSo, the electric field is given by, E \$= \frac{V}{d} = \frac{10}{5} \times 100 =\$ 200Direction of Electric field is from higher potential surface towards the lower potential surface . Hence , an Obtuse angle .

Question 1

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 2

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 3

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 4

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 5

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 6

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 7

Figure shows three spherical and equipotential surfaces $A$, $B$ and $C$ around a point charge $q$. The potential difference ${ V }_{ A }-{ V }_{ B }={ V }_{ B }-{ V }_{ C }$. If ${ t }_{ 1 }$ and ${ t }_{ 2 }$ be the distances between them, then :

Accepted Answer

The potential around a point charge is $V(r) = \displaystyle \frac{q}{4\pi\epsilon_0r}$Let $a,b,c$ be the radii of $A,B,C$ respectively.Thus, $t_1 = b-a$ and $t_2 = c-b$$V_A - V_B = \displaystyle \frac{q}{4\pi\epsilon_0a} - \frac{q}{4\pi\epsilon_0b} = \frac{q\ t_1}{4\pi\epsilon_0ab}$$V_B-V_C = \displaystyle \frac{q}{4\pi\epsilon_0b} - \frac{q}{4\pi\epsilon_0c} = \frac{q\ t_2}{4\pi\epsilon_0bc}$But, $V_A - V_B = V_B - V_C$$\Rightarrow \displaystyle \frac{V_A - V_B}{V_B - V_C} = 1\Rightarrow \frac{t_1}{ab}\times \frac{bc}{t_2}=1\Rightarrow\frac{t_1}{t_2} = \frac{a}{c}$As $a<c$, we have $t_1 < t_2$

Question 8

If a unit charge is taken from one point to another over an equipotential surface, then work done on the charge is :

Accepted Answer

Equipotential surface has same potential at all points on the surface.Thus potential difference between two points is zero i.e. $\Delta V = 0$ Work done in moving a charge over an equipotential surface $W = q\Delta V$$\implies$ $W = 0$

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

Equipotential surfaces are shown in the figure. Find the magnitude and the direction of the resultant electric field.

Accepted Answer

Since, each of the equipotential surfaces are parallel to each other,the perpendicular distance between them is given byd $= 10 sin 30 =$ 5Now, the potential gradient between 2 consecutive planes is 10VSo, the electric field is given by, E $= \frac{V}{d} = \frac{10}{5} \times 100 =$ 200Direction of Electric field is from higher potential surface towards the lower potential surface . Hence , an Obtuse angle .

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Equipotential Lines And Equipotential Surfaces

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Equipotential surface and its properties

Equipotential Surface of Various Charge System

In a region of constant potential:

Define equipotential surface.

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 ?

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

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

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 :

Figure shows three spherical and equipotential surfaces $A$ , $B$ and $C$ around a point charge $q$ . The potential difference $V_{A} - V_{B} = V_{B} - V_{C}$ . If $t_{1}$ and $t_{2}$ be the distances between them, then :

If a unit charge is taken from one point to another over an equipotential surface, then work done on the charge is :

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

Equipotential surfaces are shown in the figure. Find the magnitude and the direction of the resultant electric field.

Revise with Concepts

Equipotential Lines And Equipotential Surfaces

Learn with Videos

Quick Summary With Stories

Equipotential surface and its properties

Equipotential Surface of Various Charge System

In a region of constant potential:

Define equipotential surface.

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 ?

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

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

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 :

Figure shows three spherical and equipotential surfaces A, B and C around a point charge q. The potential difference VA​−VB​=VB​−VC​. If t1​ and t2​ be the distances between them, then :

If a unit charge is taken from one point to another over an equipotential surface, then work done on the charge is :

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

Equipotential surfaces are shown in the figure. Find the magnitude and the direction of the resultant electric field.

Figure shows three spherical and equipotential surfaces $A$ , $B$ and $C$ around a point charge $q$ . The potential difference $V_{A} - V_{B} = V_{B} - V_{C}$ . If $t_{1}$ and $t_{2}$ be the distances between them, then :