Question

The drawings show three charges that have the same magnitude but may have different signs. In all cases the distance d between the charges is the same. The magnitude of the charges is |q|= 1.6 µC, and the distance between them is d = 2.6 mm. Determine the magnitude of the net force on charge 2 for each of the three drawings.

Answer 1

Answer:

a)$F = 6816.5680 N$

b) $F' = 0 N$

c)$F''=28195.5N$

Explanation:

Magnitude of charges $M=1.6\muC$

Distance $d=2.6$

Generally the equation for  Net Force is mathematically given by

For First Drawing

$F =\frac{ k q^2}{d^2}+\frac{k q^2}{d^2}$  

$F =\frac{2* 9*10^9* (1.6*10^-6)^2}{ (2.6*10^-3^2}$

$F = 6816.5680 N$

For second Drawing

$F' =\frac{ k q^2}{d^2 }-\frac{ k q^2}{ d^2}$  

$F' = 0 N$

For Third Drawing

$F'' =\frac{ k q^2}{d^2} * \sqrt (2)$

$F'' = 9*10^9* (6.4*10^-6)^2 * \frac{\sqrt(2)}{(4.3*10^-3)^2}$

$F''=28195.5N$