Electric Field Intensity Assignment Help

Electrostatics - Electric Field Intensity

Electric Field Intensity

The electric field intensity at any point is the strength of electric field at that point. It is defined as the force experienced by unit positive charge placed at that point. If f is the force acting on a small test charge + qo at any point then electric field intensity at this point is given by 

® = f® / q0

The S.I. unit of electric field intensity is Newton per coulomb (N/C)

Electric field intensity is obviously a vector; the direction of E is the same as the direction of F E is along the direction in which the test charge +qo would tend to move of free to do so.

For a positive source charge, the electric field will be directed radically outwards from the charge. If the source charge. Therefore to minimise this effect and ultimately remove it, we rewrite electric field intensity at r as 

E (r) = (limit /q0) --> (OF ®/q0

As the magnitude of test charge qo decreases, electric field intensity E (r) is defined more and more accurately. On account of discrete nature of charge, the minimum possible value of charge is 1. 6 x 10-19coulomb, which is the unit charge, it cannot be zero. 

However on the macroscopic scale it is as good as taking the limit q0 --> 0. if the small test charge is positive the measured value of electric intensity will be somewhat less than the actual value of electric intensity, however if the small test charge is negative the measured value of electric intensity will be somewhat more than the actual value of electric intensity 

Note electric field E due to source charge Q does not depend upon test charge qo this is because E ∝ q0 so that the ratio F/qo does not depend on q0 all over the space, we shall get different values of E due to source charge Q. thus electric field due to source charge shall depend on space co-ordinate. 

Table. Values of some electric fields

Field location Value of E (N/C) or V/m
At the surface of uranium nucleus 3 x 1021
Within a hydrogen atom 5 x 1011
Electric break down of dry air 3 x 106
Near a charged comb 103
In a copper wire of house hold circuits 10-2

Suppose we have to calculate electric field intensity at any point p due to a point charge q at O where OP = r imangine a small positive test sharge qo at p. according to coulomb’s law force at p is 

F = (1/4π∈0) Q (q0 /r2) r

Where r is unit vector directed form Q towards q0

As E = F / q

∴ E = (1 / 4π0) (Q/ r2) r

 Now suppose the point charge + Q is situated at A, where OA = r1 we have to calculate electric field intensity (E) at B, where OB = r

According to coulomb’s law force on a small test charge (+ q0at B is 

F = (1 / 4π∈0)Q (q0 /r2) r 212 e = (1 / 4π0) Q (q0 /r 312 r12)

F = {(1 / 4π∈0)Q (q0 /|r2 – r1|3)(r2 – r1)}

Where AB = r12 = r2 – r1  

As E = F / q0

∴ E {[Q / (4π∈0 )] [Q / (|r2 – r1|3) (r2 –r1)]}

Obviously it is along AB produced.

In general electric field intensity (E) at position (r) due to point charge Q at (r0) is 

E = {(Q / 4π∈0) |r – r0|3 (r – r0)}

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