Four charged particles are at the corners of a square

Posted By Admin @ September 03, 2022


Four charged particles are placed so that each particle is at the corner of a square. The sides of the square are 15 cm. The charge at the upper left corner is +3.0 ?C, the charge at the upper right corner is ?6.0 ?C, the charge at the lower left corner is ?2.4 ?C, and the charge at the lower right corner is ?9.0 ?C. A. What is the net electric force on the + 3.0 ?C charge? B. What is the net electric force on the ? 6.0 ?C charge? C. What is the net electric force on the ? 9.0 ?C charge?


(i) Part 1

Net force on 3 uC charge

Force due to - 6 uC

F_1 = \frac{(9\times 10^9)(3\times 10^{-6})(6\times 10^{-6})}{0.15^2}

F_1 = 7.2 N

Force due to -2.4 uC charge

F_2 = \frac{(9\times 10^9)(3\times 10^{-6})(2.4\times 10^{-6})}{0.15^2}

F_2= 2.88 N

Force due to 9 uC charge

F_3 = \frac{(9\times 10^9)(3\times 10^{-6})(9\times 10^{-6})}{2(0.15)^2}

F_3 = 5.4 N

Now net force is given as

F_x = F_1 + F_3cos45

F_x = 11.02 N

F_y = F_2 + F_3sin45

F_y = 6.70 N

F_{net} = \sqrt{F_x^2 + F_y^2}

F_{net} = 12.9 N

Part 2

Net force on -6 uC charge

Force due to 3 uC

F_1 = \frac{(9\times 10^9)(3\times 10^{-6})(6\times 10^{-6})}{0.15^2}

F_1 = 7.2 N

Force due to 9 uC charge

F_2 = \frac{(9\times 10^9)(6\times 10^{-6})(9\times 10^{-6})}{0.15^2}

F_2= 21.6 N

Force due to -2.4 uC charge

F_3 = \frac{(9\times 10^9)(6\times 10^{-6})(2.4\times 10^{-6})}{2(0.15)^2}

F_3 = 2.88 N

Now net force is given as

F_x = F_1 - F_3cos45

F_x = 5.16 N

F_y = F_2 + F_3sin45

F_y = 23.6 N

F_{net} = \sqrt{F_x^2 + F_y^2}

F_{net} = 24.2 N


Net force on 9 uC charge

Force due to - 6 uC

F_1 = \frac{(9\times 10^9)(9\times 10^{-6})(6\times 10^{-6})}{0.15^2}

F_1 = 21.6 N

Force due to -2.4 uC charge

F_2 = \frac{(9\times 10^9)(9\times 10^{-6})(2.4\times 10^{-6})}{0.15^2}

F_2= 8.64 N

Force due to 3 uC charge

F_3 = \frac{(9\times 10^9)(3\times 10^{-6})(9\times 10^{-6})}{2(0.15)^2}

F_3 = 5.4 N

Now net force is given as

F_x = F_2 - F_3cos45

F_x = 4.82 N

F_y = F_1 - F_3sin45

F_y = 17.8 N

F_{net} = \sqrt{F_x^2 + F_y^2}

F_{net} = 18.4 N

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