state and explain the principle of superposition in electrostatics ?
Chapter1
Principle of Superposition of Charges
According to principle of superposition force on any charge due to a number of other charges is the vector sum of all the forces on that charge due to the other charges, taken one at a time. The individual forces are unaffected due to the presence of other charges.
To better understand the concept, consider a system of three charges q_{1}, q_{2} and q_{3} as shown in Fig.
The force on one charge, say q_{1 }due to two other charges q_{2}, q_{3} can therefore be obtained by performing a vector addition of the forces due to each one of these charges. Thus, if the force on q1 due to q2 is denoted by and is given by following even though other charges are present
In the same way, the force on q_{1 }due to q_{3}, denoted by and is given by
Which again is the Coulomb force on q_{1} due to q_{3} , even though other charge q_{2} is present. Thus the total force on q_{1} due to the two charges q_{2 }and q_{3} is given as
The above calculation of force can be generalised to a system of charges more than three and according to the principle of superposition in a system of charges q_{1} , q_{2} , …, q_{n} , the force on q_{1} due to q_{2} is the same as given by Coulomb’s law, i.e., it is unaffected by the presence of the other charges q_{3} , q_{4} , …, q_{n} . The total force F1 on the charge q_{1}, due to all other charges, is then given by the vector sum of the forces,, …,
The vector sum is obtained as usual by the parallelogram law of addition of vectors.
Electric Field:Electric field is the environment created by an electric charge (source charge) in the space around it, such that if any other electric charges (test charges)is present in this space, it will come to know of its presence and exert a force on it.
Figure: Electric Field due charge Q
Electric Filed Intensity: Intensity of electric field at a point is measured by force being experienced by a charge particle at that point.Hence if intensity of electric field at a location is E and a charge ‘q’ is placed, then force experienced by this charges is
…………………….(1)
Or
If q=1 unit then= , hence electric field is force experienced by unit charge is Called electric field. Electrical field is vector quantity and its direction is same as that of force experience by test charge.

Force experienced by test charge (q) is
Or ………………………(2)
From equation (1) and (2)
Or ………………(3)
Here Q is source charge and q is test charge if test is given then electric field intensity is measured by equation (1) and if Source charge is given then electric field intensity is measured by equation (2). The unit of electric field intensity is and dimension is that is . Electric field is directed towards source charge is negative and if electric field directed towards away from source charge is positive electric field.
Figure: Direction of Electric Field Due to Negative Charge  Figure: Direction of Electric Field Due to Positive Charge 
Distribution of Charge: Electric charge on a body may be concentrated at a point, then it is called a ‘point charge’. If it is distributed all over, then it is called distribution of charge. Depending on shape of it is given different names
 Point Charge: If any charge is concentrate at a point then charge is called electric charge.
The electric field due to point charge is a radial field and at the distance of r form the point charge then the intensity of electric charge is
 Line Charge:
When charge is evenly distributed over a length and inthis case the electric field intensity a point away from this is different because the direction of electric field from different point of linear charge is different at that point as shown in figure.
Figure: Electric Field Intensity at a point away from charge due to line Charge
According to above diagram the vertical component of electric field intensity at point P that is R distance from the linear charge will canceled out due to opposite direction. So only horizontal component will give electric field intensity on point P due to linear charge.Hencethe electric field intensity at a point P away from charge is less than electric field intensity due to point charge.
To calculate the electric field intensity at point P that is R distance away from the charge assume linear charge density λ. Which has relation , Where ‘Q’ is charge distributed over a long conductor of length ‘L’. Also we consider infinitesimal small element on the line charge and this very small element can be considered a point charge for point P. For point charge we can apply Coulumb’s law. The charge due to element is and electric field intensity at point is
……………………………..(1)
Total electric field due to line charge is
…………………………….(2)
Areal distribution : In this case charge is evenly distributed over a surface area S.In this also the vertical component of electric field will canceled out due to opposite direction only horizontal component will give the electric field intensity at a point P away from surface area. So to find out the electric field at a point P away from the surface assume a special quantity areal charge density (Sigma) which has relation , where is total charge on surface area and S is surface area.
Figure: Electric Field Intensity at a point away from charge due to line Charge.
Electric field intensity due to very small elemental area at point P that is r distance from the surface area is
……………………(3)
The total charge due to surface charge on point P is
………………………………(4)
Volumetric Distribution:In this case charge is evenly distributed with in volumeV. So to find out the electric field at a point P away from the volume charge assume a special quantity volumetric charge density (Rho) which has relation , where Q is total charge on total volume and V is total volume. And charge carried by very small volume is . The electric field intensity at point P that is r distance away from charge, due to elementary volume is
……………………………. (5)
Total electric field intensity at point P is
…………………………..(6)
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