(a) The situation is represented in the given figure. O is the mid-point of line AB.
Distance between the two charges, AB = 20 cm
∴ AO = OB = 10 cm
Net electric field at point O = E
Electric field at point O caused by +3μC charge,
Where, 0 = Permittivity of free space and 1/4πεo=9×109 Nm2C−2
Therefore,
Magnitude of electric field at point O caused by −3μC charge,
[ Since the magnitudes of E1 and E2 are equal and in the same direction ]
Therefore, the electric field at mid-point O is 5.4 × 106 N C−1 along OB.
(b) A test charge of amount 1.5 × 10−9 C is placed at mid – point O.
q = 1.5 × 10−9 C
Force experienced by the test charge = F
The force is directed along line OA. This is because the negative test charge is repelled by the charge placed at point B but attracted towards point A. Therefore, the force experienced by the test charge is 8.1 × 10−3 N along OA.
