Two point charges of 10 esu and 40 esu are located at points A and B seperated by a distance 4 cm. Find the electric field intensity at a point halfway between the charges.

Explanation

Since the charges are 4 cm apart, each of them away from the midpoint (O) joining them [Fig. 2.34] Electric field intensity at O due to the charge 10 esu,
`E_(1)=10/((2)^(2))=2.5` dyn/esu, along `vec(OB)`

Electric field intensity at O due to the charge 40 esu,
`E_(2)=40/((2^(2))=10` dyn/esu , along `vec(OA)`
Since `vec(E_(1))` and `vec(E_(2))` are oppositely directed, the resultant and intensity at O,
`E=E_(2)-E_(1)=10-2.5=7.5` dyn/esu, along `vec(OA)`