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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=14πε0⋅3×10−6(OA)2=14πε0⋅3×10−6(10×10−2)2NC−1 along OB
Where, Eo = Permittivity of free space and 14πε0=9×109Nm2C−2
Therefore,
The magnitude of the electric field at point O caused by - 3μC charge,
E2=|14πε0⋅−3×10−6(OB)2|=14πε0⋅3×10−6(10×10−2)2NC−1 along OB∴E=E1+E2=2×14πε0⋅3×10−6(10×10−2)2NC−1 along OB∴E=2×9×109×3×10−6(10×10−2)2NC−1=5.4×106NC−1 along OB
Therefore, the electric field at mid-point O is 5.4×106NC−1 along OB.
 
(b) A test charge of amount 1.5×10−9C is placed at midpoint O.
q=1.5×10−9C
Force experienced by the test charge = F
∴F=qE=1.5×10−9×5.4×106=8.1×10−3N
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 8.1×10−3 N along OA.

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