Q 1.8) Two point charges qA = 3 µC and qB= –3 µC are located 20 cm apart in a vacuum.(i) What is the electric field at the midpoint O of the line AB joining the two charges?(ii) 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?

 
 
 
(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
The net electric field at point O = E

The electric field at point O caused by + 3μC charge,
 
E1=14πε03×106(OA)2=14πε03×106(10×102)2NC1 along OB
Where, Eo = Permittivity of free space and 14πε0=9×109Nm2C2
Therefore,
The magnitude of the electric field at point O caused by - 3μC charge,
E2=|14πε03×106(OB)2|=14πε03×106(10×102)2NC1 along OBE=E1+E2=2×14πε03×106(10×102)2NC1 along OBE=2×9×109×3×106(10×102)2NC1=5.4×106NC1 along OB
Therefore, the electric field at mid-point O is 5.4×106NC1 along OB.
 
(b) A test charge of amount 1.5×109C is placed at midpoint O.
q=1.5×109C
Force experienced by the test charge = F
F=qE=1.5×109×5.4×106=8.1×103N
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 to point A.
Therefore, the force experienced by the test charge is 8.1×103 N along OA.