Question 8

Two point charges q_{A} = 3 μC and q_{B} = −3 μC are located 20 cm apart in vacuum.

(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?

Answer

(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,

*E*_{1} = along OB

Where,

= Permittivity of free space

Magnitude of electric field at point O caused by −3μC charge,

along OB

= 5.4 × 106 N/C along OB

Therefore, the electric field at mid-point O is 5.4 × 10^{6} 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 × 10^{6}

= 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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Prince
2019-05-14 11:41:34

Why did you take 10^-2 on multiplying with 4^2?

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