Question #190230

Three point charges as shown in the accompanying figure are fixed in place in a right triangle. Determine the electric force on the -0.60 µC charge due to the other two charges.

Expert's answer

assuming that the two charges for q2 = + 3"\\mu C"

q3 = - 4"\\mu"C

to find the resultant force that acts on the q1= - 6 "\\mu" C we need to find resultant forces of pairs that is ;

magnitude of force on q1 by q2 that that will be called F2

F2 = "k\\times \\frac{q_1q_2}{d^2} ="

"( \\frac{6\\times 10^{-6}\\times 3\\times10^{-6}}{0.03^2}) \\times 9.0 \\times 10^9=" 180 N

magnitude of force on q1 from q3 = F3

F3= "k\\times \\frac{q_1q_3}{d^2} ="

"( \\frac{6\\times 10^{-6}\\times 4\\times10^{-6}}{0.04^2}) \\times 9.0 \\times 10^9" = 135 N

to find the resultant force Fr Acting on Charge Q1 we apply Pythagoras rule for right angles.

Fr= "\\sqrt{ F_2^2+ F_3^2} = \\sqrt{ 180^2+ 135^2}"

Fr= 225 N

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