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JEE Mains

## Coulomb's Law

Coulomb measured the force between two point charges and found that it varied inversely as the square of the distance between the charges and was directly proportional to the product of the magnitude of the two charges and acted along the line joining the two charges. Thus, if two point charges q1, q2 are separated by a distance r in vacuum, the magnitude of the force (F) between them is given by
(a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is C? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
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## Relation between electric field lines and flux across a surface

The field lines carry information about the direction of electric field at different points in space. Having drawn a certain set of field lines, the relative density (i.e., closeness) of the field lines at different points indicates the relative strength of the electric field at those points. The field lines are crowded where the field is strong and are spaced apart where it is weak.
While electric flux is the measure of the flow of the electric field through a given area. It is proportional to the number of electric field lines passing normally through a  perpendicular surface.

Figure shows the field produced by two point charges +q and -q of equal magnitude but opposite signs (an electric dipole). Find the electric flux through each of the closed surfaces A, B, C and D.
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## Density of electric field lines

Magnitude of electric field strength is higher where density of electric field lines in space is more. The attached figure shows the electric field lines due to a positive point charge. Its field lines are directed radially outwards from the point charge and the dotted lines show the locus of points with same magnitude of electric field.
Since, for a point charge, it is larger near the charge. This behaviour is consistent in the electric field lines of point charge.
The figure shows the electric field lines around a conductor.
(a) At which of the four points is the field strongest?
(b) At which point is there a negative charge?
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## Electric field due to a continuous charge distribution

Example:
A rod of length with uniform charge per unit length is placed at a distance from origin along the x-axis. A similar rod is placed at the same distance along Y-axis. Determine the magnitude of net electric field intensity at the origin.
Solution:
Let's first find Electric field due to the rod on x-axis, say
Consider a small length element , distance away from origin.
Field  due to this
Integrating from to
Gives along -x direction
Now, is along the -y direction and same in magnitude.
Thus,

A given charge situated at a certain distance from an electric dipole in the end on position, experiences a force F. If the distance of the charge is doubled, the force acting on the charge will be :
A
F/8
B
F/4
C
F/2
D
2F
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## Electric field due to a charged ring along the axis

where
is the radius of the ring
is the charge density
is the distance from the centre of the ring along the axis
The linear charge density of four quadrants of a ring of radius R is as shown. Electric field at the centre will be :
A
Zero
B
C
D
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## Electric field due to non-uniformly charged sphere

Example:
A non-uniformly charged sphere of radius has a charge density where is constant and is the distance from the center of the sphere. Find the electric field inside and outside the sphere.
Solution:
Electric field lines will be directed radially outwards as the charge is distributed radially uniformly.
For ,
Charge enclosed in an infinitesimally small spherical shell of thickness at a distance is:

Charge enclosed upto distance is:

Applying gauss law at a distance

For ,
Charge enclosed is:

Electric field outside is hence given by:

A solid ball of radius has a charge density given by for . The electric field outside the ball is:
A
B
C
D
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## SHM in electrostatic force

Example:
A particle of mass and charge is at rest at the mid point between two other fixed similar charges each of magnitude placed a distance apart. The system is collinear as shown in figure. The particle is now displaced by a small amount along the line joining the two charges and is left to itself. It will now oscillate about the mean position with what time period?( permittivity of free space)
Solution:
Restoring force on displacement of

Acceleration
or

A charge +q is placed at and another charge is placed at . A charge is placed at the origin. If it is slightly displaced along y axis. Then :
A
it will move away
B
it will oscillate but will not execute SHM
C
it will execute SHM
D
it will stand at the displaced position
Assuming mass m of the charged particle, find time period of oscillation
A
B
C
D
none of these
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## Electric field lines of simple configurations

content Text

Figure shows lines of force for a system of two point charges. The possible choice for the charges is:

A
B
C
D
The figure is a plot of lines of force due to two charges and .
Find out the sign of charges
A
both negative
B
Upper positive and lower negative
C
both positive
D
upper negative and lower positive
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## Electric field due to a uniformly charged disc

where and is the surface charge density
A thin uniformly charged disc can be imagined as a
A
discrete collection of concentric rings
B
continuous collection of concentric rings
C
discrete collection of chords
D
continuous collection of chords
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## Define and calculate electric flux density and identify its unit

The Electric Flux Density (D) is related to the Electric Field (E) by:
-----[Equation 1]
In Equation [1], is the permittivity of the medium (material) where we are measuring the fields.
The Electric Field is equal to the force per unit charge:
------[Equation 2]
Then the Electric Flux Density is:
--------[Equation 3]
From Equation [3], the Electric Flux Density is very similar to the Electric Field, but does not depend on the material in which we are measuring (that is, it does not depend on the permittivity. Note that the field is a vector field, which means that at every point in space it has a magnitude and direction.
The Electric Flux Density has units of Coulombs per meter squared [].
A point charge placed on the axis of a uniformly charged disc experiences a force F due to the disc. If the surface charge density on the disc is , the electric flux through the disc due to the point charge will be :
A
B
C
D
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## Application of Gauss Law in a given non uniform electric field

Example:
An electric field given by  passes Gaussian cube of side 1m placed with one corner is at origin such that its sides represents x, y and z axes. Find the magnitude of net charge enclosed within the cube.
Solution:
The flux passing through the surfaces parallel to x-y plane is zero as their is no z component of E.
The flux passing through the surfaces parallel to y-z plane is zero as the x component of E is a constant. Flux going in goes out, so no net flux.
The flux passing through the surface parallel to x-z plane is given by,
and

Magnitude of charge enclosed
A charge is placed at the corner of a cube of side . The electric flux passing through the cube is :
A
B
C
D
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## Redistribution of charges for arrangement of parallel plates

Total charge on A is and on B is which remains constant by law of conservation of charge.
From gauss's law, total charge on the inner regions is zero. Hence, equal and opposite charges are induced on the inner surface.
Net electric field at P is zero.

Two large, parallel conducting plates X and Y, kept close to each other, are given charges and . The four surfaces of the plates are A, B, C and D, as shown in figure. Then :
This question has multiple correct options
A
The charge on A is .
B
The charge on B is .
C
The charge on C is -.
D
The charge on D is
Five identical plates are connected across a battery as in figure. If the charge on plate be , then the charges on the plates and are:
A
B
C
D
none of the above.
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## Force on electric dipole

A small electric dipole having dipole moment is placed along X-axis, as shown in the figure. A semi-infinite uniformly charged di-electric thin rod  placed along x axis, with one end coinciding with origin. If linear charge density rod is and distance of dipole from rod is , then calculate the electric force acting on dipole.
Electric field at A due to dipole,              where

Force at A due to dipole,

Total force exerted due to dipole on the rod,

Now force exerted by rod on dipole is equal and opposite to  (Newton's law)
An electric dipole is placed perpendicular to an infinite line of charge at some distance as shown in figure. Identify the correct statement.
A
The dipole is attracted towards the line charge
B
The dipole is repelled away from the line charge
C
The dipole does not experience a force
D
The dipole experiences a force as well as a torque
An electric dipole is placed in an electric field increasing at the rate of per unit distance along the x-axis. The dipole may undergo :
A
rotational motion only
B
rotational motion and translational motion along z-axis
C
rotational motion and translational motion along x-axis
D
only translational motion along x-axis