Question -
Answer -
(a) Thesituation 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
Electricfield at point O caused by +3μC charge,
E1 =
along OB
Where,
Permittivity of free space
Magnitude of electric field atpoint O caused by −3μC charge,
E2 = 
= 
along OB
= 5.4 × 106 N/Calong OB
Therefore, the electric field atmid-point O is 5.4 × 106 N C−1 along OB.
(b) A testcharge of amount 1.5 × 10−9 C is placed at mid-point O.
q =1.5 × 10−9 C
Force experienced by the testcharge = F
∴F = qE
= 1.5 × 10−9 ×5.4 × 106
= 8.1 × 10−3 N
The force is directed along lineOA. This is because the negative test charge is repelled by the charge placedat point B but attracted towards point A.
Therefore, the force experiencedby the test charge is 8.1 × 10−3 N along OA.