4,555.*: Cand Haris Three charged me to conducting surface de the day as der - 2.00m...

Question

Question

Answer 1

Answer #1

Electric field due to q1 will be to the right

Electric field due to q2 will be to the left

Electric field due to q3 will be to the right

_______________________

Therefore,

E = + 9e9 * 5.55e-6 / 0.02² - 9e9 * 1.31e-6 / 0.01² + 9e9 * 2.49e-6 / 0.03²

E = 1.2487e8 - 1.179e8 + 2.49e7

E = 3.187e7 N/C

______________________

and

force

F = qE

F = - 3.56e-6 * 3.187e7

F = 113.45 N ( to the left)

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Answer 2

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