Answer to Question #190230 in Civil and Environmental Engineering for John Prats

Question

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.

Answer 1

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