Question

6) A 5nC charge is placed at the origin of our coordinate system. There are three other charges; 5C located 3m in the positive i direction, 2C located 2m in the negative i direction, and -4C located 5m in the positive j direction. What is the net force acting on the 5nC charge? Remember that force is a vector, so your answer must include magnitude and direction. Component form is acceptable.

7) There are two charges, one +2C, the other -1C, separated by a distance of 2m. Find the point(s) where electric field is zero.

Answer 1

"electric field due to 2c ^5c E_25 = k [-5/3^2 + 2/2^2] i E_25 = -5 times 10^2 i electric field due to -4c E_4 = k 4/5^2 = 1.44 times 10^9 j Net electric field E = E_4 + E_5 = -5 times 10^2 i + 1.44 times 10^9 j Force experienced F = 9E = 5 times 10^-9 (-5 times 10^2 i + 1.44 times 10^9 j) F = -2.5 i + 7.2 j rightarrow component form

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