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.

Respuesta :

Answer:

[tex]3.13\times 10^{-5} C[/tex]

Explanation:

We are given that

[tex]q_1=26\mu C=26\times 10^{-6} C[/tex]

[tex]1\mu C=10^{-6} C[/tex]

[tex]y_1=0.21 m[/tex]

[tex]q=8.1\mu C=8.1\times 10^{-6} C[/tex]

[tex]y_2=0.39 m[/tex]

F=28 N

We have to find the magnitude of q2.

We know that

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

Where [tex]k=9\times 10^9[/tex]

Using the formula

Force on charge q due to charge q1

[tex]F_1=\frac{9\times 10^9\times 26\times 10^{-6}\times 8.1\times 10^{-6}}{(0.21)^2}[/tex]

[tex]F_1=42.98 N[/tex]

Force on charge q due to point charge q2

[tex]F_2=\frac{9\times 10^9\times q_2\times 8.1\times 10^{-6}}{(0.39)^2}[/tex]

[tex]F_2=4.79\times 10^5 q_2[/tex]

[tex]F=F_1-F_2[/tex]

[tex]28=42.98-4.79\times 10^5q_2[/tex]

[tex]4.79\times 10^5q_2=42.98-28=14.98[/tex]

[tex]q_2=\frac{14.98}{4.79\times 10^5}=3.13\times 10^{-5} C[/tex]

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