Question

As shown in the diagram, threee equal charges are spaced evenly in a row. The magnitude of each charge is +2e, and the distance between two adjacent charges is 1.50 nm. Then the central charge is displaced 0.350 nm to the right while the other two charges are held in place. After the displacement, what is the magnitude and direction of the net force that the outer two charges exert on the central charge?
4.27*10^-10 to the right

4.27*10^-10 to the left

3.20*10^-10 to the left

3.20*10^-10 to the right

Answer 1

Answer:

4.27*10^(-10) to the left

Explanation:

Force exerted by right charge

$F_{right} = \frac{k*(2e)*(2e)}{((1.5 - 0.35)*10^(-9))^2} \\\\F_{right} = \frac{8.99*10^9 * 4 * (1.6*10^(-19))^2}{((1.5 - 0.35)*10^(-9))^2} \\\\F_{right} = 6.96*10^(-10) N$

Force exerted by left charge

$F_{left} = \frac{k*(2e)*(2e)}{((1.5 + 0.35)*10^(-9))^2} \\\\F_{left} = \frac{8.99*10^9 * 4 * (1.6*10^(-19))^2}{((1.5 + 0.35)*10^(-9))^2} \\\\F_{left} = 2.689*10^(-10) N$

Resultant Force

$F_{res} = F_{right} - F_{left}\\F_{res} = 6.96*10^(-10) - 2.689*10^(-10)\\\\F_{res} = 4.271 * 10 ^(-10) N$

Hence, right charge exerts more force than left so central experience the above force in left direction.

Answer 2

Answer: 4.27 x 10^-10 N to the left

Explanation: I just took this quiz