What is the magnitude of the electric field 27.6 cm directly above an isolated 3.50×10−5 C...

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Question

What is the magnitude of the electric field 27.6 cm directly above an isolated 3.50×10−5 C charge? What is the direction of the electric field?

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Answer 1

Answer #1

Electric field of a charge Q at a distance r from it is given by

E = kQ/r²

E = (9*10⁹) * (3.50*10^-5C) / (27.6*10^-2m)²

E = 4.135*10⁶ N/C

electric field is directed vertically upwards.

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Answer 2

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