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(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 × 10−9 C is placed at this point, what is the force experienced by the test charge?
Asked by Vivekchettri Last Modified
Answer 
Apoorva Purohit
(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,
E1 = 
along OB
Where,
= Permittivity of free space
Magnitude of electric field at point O caused by −3μC charge,
E2 = 
= 
along OB
= 5.4 × 106 N/C along OB
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
∴F = qE
= 1.5 × 10−9 × 5.4 × 106
= 8.1 × 10−3 N
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.
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