Electric field
The region or space around a given charge in which some other charge experience an electrostatic force on it , is called the electric field of the (given) charge.
Theoretically, the electric field of a given charge is extended upto infinity.
The Intensity of Electric Field (Measure of Electric Field)
“The intensity of electric field at a point is equal to the electrostatic force experienced by unit positive charge placed at that point”.
Electric field (intensity) is a vector quantity and it is directed along the direction of the force experienced by the positive test charge at that point.
Electric field due to a positive charge is always away from the charge and that due to a negative charge is always towards the charges.
Unit of [E] : newton/coulomb = volt/metre
Dimensions of [E] : = [MLT–3 A–1 ]
Question 1. A positively charged ball hangs from a long silk thread. We wish to measure E at a point in the same horizontal plane as that of the hanging charge. To do so, we put a positive test charge q0 at the point and measure F/q0. Will F/q0 be less than, equal to, or greater than E at the point in question?
Solution.
When we try to measure the electric field at point P then, after placing the test charge at P, it repels the source charge (suspended charge) and the measured value of electric field Emeasured = F/qo = will be less than the actual value Electric field E.
Electric Field Due to A Point Charge:
Electric Field intensity at point ‘P’, at distance r from a point charge ‘q’ is
Electric Field (basic Concept)
Example. Calculate the electric field intensity which would be just sufficient to balance the weight of a particle of charge –10m C and mass 10 mg.
Solution. As force on a charge q in an electric field is
F = q.E
So according to given problem
i.e., |q|E = mg
i.e., E = F/|q|= 10 N/C., in downward direction.
Example. Electrostatic force experienced by –3μ C charge placed at point P due to a point charge system S as shown in figure is 
(i) Find out electric field intensity at point P due to S.
(ii) If now 2 μC charge is placed and –3 μC is removed at point P then force experienced by it will be.
Solution. (i)
(ii) Since the source charges are not disturbed the electric field intensity at ‘P’ will remain same.
Properties of electric field intensity ‘E’:
(i) It is a vector quantity. Its direction is the same as the force experienced by positive charge.
(ii) Electric field due to positive charge is always away from it while due to negative charge always towards it.
(iii) Its S.I. unit is Newton/Coulomb.
(iv) Its dimensional formula is [MLT–3A–1]
(v) Electric force on a charge q placed in a region of electric field at a point where the electric field intensity is E , is given by F = q.E.
Electric force on point charge is in the same direction as that of electric field, if the point charge is positive and force is in opposite direction if it is a negative charge.
(vi) It obeys the superposition principle, that is, the field intensity at a point due to a point charge distribution is vector sum of the field intensities due to individual point charges.
Example. Two point charges 2 μC and – 2 μC are placed at point A and B as shown in figure. Find out electric field intensity at points C and D. [All the distances are measured in metre].
Solution. Electric field at point C
(EA, EB are magnitudes only and arrows respresent directions)
Electric field due to positive charge is away from it while due to negative charge it is towards the charge. It is it is clear that EB > EA .
Electric field at point D :
SInce magnitude of charges are same and also AD = BD
So EA = EB
Verticle components of and cancel each other while horizontal components are in the same direction.
