Question

1.) Four equal magnitude point charges (3 uC) are placed at the corners of a square that is 4 cm on a side. Two diagonally opposite each other, are positive and the other two are negative. Determine the force on either negative charge.

2.) Charges of +2,+3, and -8 uC are placed at the vertices of an equilateral triangle of side 10 cm. Calculate the magnitude of the force acting on the -8 uC due to the other two charges

Answer 1

Answer on Question #74185-Physics-Electromagnetism

1.) Four equal magnitude point charges (3 uC) are placed at the corners of a square that is 4 cm on a side. Two diagonally opposite each other, are positive and the other two are negative. Determine the force on either negative charge.

Solution

$$F_1 = F_2 = \frac{k Q^2}{a^2} = \frac{(9 \cdot 10^9)(3 \cdot 10^{-6})^2}{(0.04)^2} = (50.625) N.$$ $$F_3 = \frac{k Q^2}{\left(\sqrt{2} a\right)^2} = \frac{(9 \cdot 10^9)(3 \cdot 10^{-6})^2}{(0.04\sqrt{2})^2} = (25.3125) N.$$ $$F_x = F_y = F_1 - \frac{F_3}{\sqrt{2}}$$

Therefore, the force on either negative charge

$$F = \sqrt{2} \left(F_1 - \frac{F_3}{\sqrt{2}}\right) = \sqrt{2} F_1 - F_3 = \sqrt{2} (50.625) - (25.3125) = 46.3 N.$$

2.) Charges of $+2, +3$, and $-8$ uC are placed at the vertices of an equilateral triangle of side $10$ cm. Calculate the magnitude of the force acting on the $-8$ uC due to the other two charges

Solution

$$F_1 = \frac{k q_1 Q}{a^2} = \frac{(9 \cdot 10^9)(2 \cdot 10^{-6})(8 \cdot 10^{-6})}{(0.1)^2} = (14.4) N.$$ $$F_2 = \frac{k q_2 Q}{a^2} = \frac{(9 \cdot 10^9)(3 \cdot 10^{-6})(8 \cdot 10^{-6})}{(0.1)^2} = (21.6) N.$$

The magnitude of the force acting on the $-8$ uC due to the other two charges is

$$F = \sqrt{F_1^2 + F_2^2 + 2 F_1 F_2 \cos \alpha}$$ $$F = \sqrt{(14.4)^2 + (21.6)^2 + 2 (14.4)(21.6) \cos 60} = 31.4 N.$$

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