Question

+Q1 = 10 μC, +Q2 = 50 μC and Q3 are separated as shown in the figure below. What is the electrostatic charge on particle 3 if the net electrostatic force on particle 2 is zero? Solution

Answer 1

$F = \frac{k * 10 *50}{r^{2}} = \frac{k*50*Q_{3} }{(r+x)^{2}}$

⇒$\frac{10}{Q_3} =\frac{r^{2} }{(r+x)^{2}}$

solve for further answer

Answer 2

Answer:

90 N

Explanation:

(K*10*10^-6*50*10^-6/0.02^2)=(K*50*10^-6*X/0.06^2)

SO YOU WILL GET

Q3= 10*0.06^2/0.02*2

THEREFORE,

Q3= 90 N