Charges are placed at the corners of a square whose sides are 2.0 m in length. The charges are +4.0 nC at (0, 0), +3.0 nC at (2.0, 0), -3.0 nC at (0, 2.0), and +4.0 nC at (2.0, 2.0). The magnitude of the force exerted on a +1.0-nC test charge placed at the square's center is

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Quizplus · Accepted Answer

The net force on the test charge is the vector sum of the forces due to each of the four charges at the corners of the square. The force due to the charge at (0, 0) is directed along the positive x-axis and has a magnitude of:$$F_1 = k\frac{q_1q}{r_1^2} = (9 imes10^9 ext{ N$\cdot$m}^2/ ext{C}^2)\frac{(4 imes10^{-9} ext{ C})(1 imes10^{-9} ext{ C})}{(1 ext{ m})^2} = 36 imes10^{-9} ext{ N}$$The force due to the charge at (2.0, 0) is directed along the negative x-axis and has a magnitude of:$$F_2 = k\frac{q_2q}{r_2^2} = (9 imes10^9 ext{ N$\cdot$m}^2/ ext{C}^2)\frac{(3 imes10^{-9} ext{ C})(1 imes10^{-9} ext{ C})}{(2 ext{ m})^2} = 6.75 imes10^{-9} ext{ N}$$The force due to the charge at (0, 2.0) is directed along the negative y-axis and has a magnitude of:$$F_3 = k\frac{q_3q}{r_3^2} = (9 imes10^9 ext{ N$\cdot$m}^2/ ext{C}^2)\frac{(-3 imes10^{-9} ext{ C})(1 imes10^{-9} ext{ C})}{(2 ext{ m})^2} = 6.75 imes10^{-9} ext{ N}$$The force due to the charge at (2.0, 2.0) is directed along the positive y-axis and has a magnitude of:$$F_4 = k\frac{q_4q}{r_4^2} = (9 imes10^9 ext{ N$\cdot$m}^2/ ext{C}^2)\frac{(4 imes10^{-9} ext{ C})(1 imes10^{-9} ext{ C})}{(2 ext{ m})^2} = 9 imes10^{-9} ext{ N}$$The net force on the test charge is:$$F = F_1 + F_2 + F_3 + F_4 = (36 imes10^{-9} ext{ N}) + (-6.75 imes10^{-9} ext{ N}) + (-6.75 imes10^{-9} ext{ N}) + (9 imes10^{-9} ext{ N}) = 27 imes10^{-9} ext{ N}$$Therefore, the magnitude of the force exerted on the +1.0-nC test charge placed at the square's center is 27 * 10^-9 N.

Charges Are Placed at the Corners of a Square Whose