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  • Two point charges q_A = 3 μC and q_B = –3 μCare located 20 cm apart in vacuum 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?

Q 1.8(b): Two point charges q_{A}=3\mu C and q_{B}=-3\mu C are located 20 cm apart in vacuum. If a negative test charge of magnitude1.5 \times 10^{-9}C is placed at this point, what is the force experienced by the test charge?

Answers (1)

best_answer

Let Q = -1.5 \times10^{-9} C

The force experienced by Q when placed at O due to the charges at A and B will be: 

F = Q \times E'

where E' is the net Electric field at the point O.

F=1.5 \times 10^{-9} C \times 5.4 \times 10^4 N/C = 8.1\times10^{-3 } N

Q being negatively charged will be attracted by positive charge at A and repelled by negative charge at B. Hence direction of force experienced by it will be in the direction of OA. 

Posted by

HARSH KANKARIA

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