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 .

Question 1

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 2

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	heta  E = \frac{{ - \Delta V}}{{\Delta r\cos 	heta }} E = \frac{{ - \left( { - 20 - 10} ight)}}{{10 	imes {{10}^{ - 2}}\cos {{120}^0}}} = \frac{{ - 10}}{{10 	imes {{10}^{ - 2}}\left( { - \sin {{30}^0}} ight)}} = \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 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	imes { 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	imes Q ight)  }{ { x }^{ 2 } }  { F }_{ 2 }=\dfrac { K\left( 9q	imes Q ight)  }{ { \left( 16-x ight)  }^{ 2 } }  Equating  { F }_{ 1 }  and  { F }_{ 2 }  we get, \dfrac { 9 }{ { x }^{ 2 } } =\dfrac { 1 }{ { \left( 16-x ight)  }^{ 2 } }  { F }_{ 1 }={ F }_{ 2 } or,   \dfrac { K\left( q	imes Q ight)  }{ { x }^{ 2 } } =\dfrac { K\left( 9q	imes Q ight)  }{ { \left( 16-x ight)  }^{ 2 } }  or,   \dfrac { 1 }{ { x }^{ 2 } } =\dfrac { 9 }{ { \left( 16-x ight)  }^{ 2 } }  or,   \dfrac { 1 }{ x } =\dfrac { 9 }{ \left( 16-x ight)  }  or,   16-x=3x or,   4x=16 or,   x=4 	herefore     (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

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 6

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 7

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 8

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 9

Three mass points each of mass  m  are placed at the vertices of an equilateral triangle of side  1  . What is the gravitational potential at the centroid of the triangle.

Accepted Answer

Let the distance of the centroid from each vertex is  r  and it can be written as,

r = \dfrac{l}{{\sqrt
3 }}

The gravitational potential at centroid due to mass  m  at A is given as,

{V_A} = \dfrac{{ -
Gm}}{r}

= \dfrac{{ - \sqrt 3
Gm}}{l}

The gravitational potentials at B and C are given as,

{V_B} = \dfrac{{ -
\sqrt 3 Gm}}{l}

{V_C} = \dfrac{{ -
\sqrt 3 Gm}}{l}

The total potential due to all the three masses is
given as,

V = {V_A} + {V_B} +
{V_C}

V = \dfrac{{ - \sqrt
3 Gm}}{l} - \dfrac{{\sqrt 3 Gm}}{l} - \dfrac{{\sqrt 3 Gm}}{l}

V = \dfrac{{ - 3\sqrt
3 Gm}}{l}

Thus, the gravitational potential at the centroid of
the triangle is  \dfrac{{ - 3\sqrt 3 Gm}}{l} .

Question 10

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.

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Equipotential surface and its properties

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Important Questions

Define equipotential surface.

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

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:

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 :

In a region of constant potential:

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

Two charges $2 μ C$ and $- 2 μ C$ 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?

Three mass points each of mass $m$ are placed at the vertices of an equilateral triangle of side $1$ . What is the gravitational potential at the centroid of the triangle.

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

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

Important Questions

Define equipotential surface.

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

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:

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 :

In a region of constant potential:

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

Two charges 2μC and −2μC 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?

Three mass points each of mass m are placed at the vertices of an equilateral triangle of side 1 . What is the gravitational potential at the centroid of the triangle.

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

Two charges $2 μ C$ and $- 2 μ C$ 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?

Three mass points each of mass $m$ are placed at the vertices of an equilateral triangle of side $1$ . What is the gravitational potential at the centroid of the triangle.