MHT-CET : Physics Entrance Exam

### MHT - CET : Physics - Electrostatics Page 1

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1.

Coulomb's Law

The magnitude of the force of attraction or repulsion between two point charges q1 and q2 separated by a distance 'r' is given by

F =

 q1 q2 4p Î0 r2

(

 1 k

)

k = dielectric constant
Î0 = permittivity of free space
Î = k Î 0 = permittivity of the medium

2.

Electric Field Intensity (E)

The electric field intensity at a point in an electric field is defined as the force exerted on a unit positive charge placed at that point.

E =

 F q

Also, E =

 1 4pk Î0

 Q r2

 3. Lines of Force 1. Electric field due to one or more charges can be picturised by drawing the lines of force. 2. A line of electric force is defined as the imaginary line or path along which a free positive charge travels when placed in an electric field. 3. These lines start from a positive charge and terminate at a negative charge. 4. They never cross each other and at any one point, an electric field can have only one direction. 5. They normally leave the surface of a charged body. 6. The magnitude of an electric field intensity at a given point is given by the number of electric lines of force passing normal to the unit area drawn around the point.

4.

Electric Flux

Electric flux through any surface in an electric field is defined as the normal component of the lines of force passing through that surface.

The lines of force in a region indicate the magnitude and the direction of electric field.

Electric field intensity E =

 Æ A

\ Æ = EA
Where is perpendicular to surface area.

If the electric field intensity is at an angle
q with the normal to a surface area ds,

 then the electric flux through the surface in normal direction is f = E A Cos q = E Cos q dS

5.

Total Normal Induction (T.N.E.I.)

The total number of lines of induction leaving a closed surface normally is called total normal induction.

 \T.N.E.I.= (dielectric constant × Î0) × (component of field intensity normal to the surface) × (area of the surface)

6.

Tubes of Force

 Consider a small surface element of a positively charged conductor. The lines of force originating from various points on this surface element form a group called a tube of force. The lines of force lie on the surface of the tube of force. The ends of the tube of force are those areas from where the lines of force originate and terminate as shown in the fig. If the surface elements are so chosen that the charge on each element is one unit, then the tube of force is called Unit tube of force or Faraday tube.

 7. Tubes of Induction The number of tubes of force originating from a charge depends on the permittivity of the medium. In order to show that number of tubes of force does not depend on the nature of the medium, Faraday induced the concept of induction. According to him, only one tube originates to form a unit positive charge, whatever be the medium surrounding this charge. Such a tube is called a tube of induction.

8.

Plane Angle and Solid Angle

The angle q in a plane is given by the ratio of the arc which subtends the angle q to the radius r.

q =

If the area ds subtends a solid angle dw at 'O' and q is the angle between the line joining O with the centre of the area and normal to the area then

dw =

 dS cosq r2

9.

Gauss's Theorem

Total normal electric induction (T.N.E.I.) through any imaginary closed surface is equal to the total quantity of charge enclosed in that surface.

 OR

The total electric flux through a closed surface is equal to 1/e times the algebraic sum of the charges enclosed by that surface.

Proof:
Suppose a charge (+Q) is situated at some point O within a closed surface S. Consider a small element of area ds on the surface S.
Take a point P on the small element which is at a distance r from the point O. The electric field intensity at point P due to charge +Q is

E =

 1 4p ke0

 Q r2

...(1)

If is unit normal to the area dS, then the electric flux over the area dS is

 dF = E. dS = E. = E cosq dS

Using (1), we get,

dF =

 1 4p ke0

 Q r2

.dScos q

\ Electric induction over the area dS

= ke0

 1 4pke0

 Q r2

. dScosq

=

 Q dS.cosq 4pr2

But ds. cosq/r2 = Solid angle dw subtended by the area dS at O.

.. Electric induction over the area dS =

 Q 4p

dw

Total normal electric induction over a

closed surface S

=

 SQ 4p

dw

=

 QS 4p

dw

But Sdw = 4p = Total solid angle subtended

by the closed surface at O

Total normal electric induction over a closed

surface S

=

 Q 4p

. 4p

= Q

 If Q1, Q2, Q3…. are the charges present within the closed surface then T.N.E.I. = Q1 + Q2 + Q3 + …….. = S Qi

10.

Application of Gauss Law

(i) The electric intensity due to a uniformly charged sphere at any point outside the sphere

E =

 q 4pÎr2

=

 q 2pkÎ0r2

=

 s kÎ0

 R r

2

 where Î = k Î0, k : dielectric constant of the medium. q : total charge r : distance of the point from the centre of the sphere. R : radius of the sphere.

Case I : Sphere having uniform surface charge density
s, and for a point very close to the charged sphere, E =

Case II : For a point inside the charged sphere, E = 0.

(ii) Electric intensity at any point outside a uniformly charged cylinder

 E =

 where q : charge per unit length r : perpendicular distance of the given point from the axis of the cylinder R : radius of the cylinder.

If the point is very close to the charged cylinder.
E =

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