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3 years ago

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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 cha

rge 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?

1 answer:

tangare [24]3 years ago

7 0

(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

$F_x = F_1 - F_3cos45$

$F_x = 5.16 N$

$F_y = F_2 + F_3sin45$

$F_y = 23.6 N$

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

$F_{net} = 24.2 N$

PART 3

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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