Question

Two point charges are fixed on the y axis a negative point charge q1-26 μC at y1 = +0.21 m and a positive point charge q2 art y2 = +0.39 m. A third point charge q = +8.1 μC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 28 N and points in the +y direction. Determine the magnitude of q2.

Answer 1

Answer:

$3.13\times 10^{-5} C$

Explanation:

We are given that

$q_1=26\mu C=26\times 10^{-6} C$

$1\mu C=10^{-6} C$

$y_1=0.21 m$

$q=8.1\mu C=8.1\times 10^{-6} C$

$y_2=0.39 m$

F=28 N

We have to find the magnitude of q2.

We know that

$F=\frac{kq_1q_2}{r^2}$

Where $k=9\times 10^9$

Using the formula

Force on charge q due to charge q1

$F_1=\frac{9\times 10^9\times 26\times 10^{-6}\times 8.1\times 10^{-6}}{(0.21)^2}$

$F_1=42.98 N$

Force on charge q due to point charge q2

$F_2=\frac{9\times 10^9\times q_2\times 8.1\times 10^{-6}}{(0.39)^2}$

$F_2=4.79\times 10^5 q_2$

$F=F_1-F_2$

$28=42.98-4.79\times 10^5q_2$

$4.79\times 10^5q_2=42.98-28=14.98$

$q_2=\frac{14.98}{4.79\times 10^5}=3.13\times 10^{-5} C$