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3 years ago

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Two point charges are fixed on the y axis: a negative point charge q1 = -34 μC at y1 = +0.18 m and a positive point charge q2 ar

t y2 = +0.33 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 31 N and points in the +y direction. Determine the magnitude of q2.

1 answer:

In-s [12.5K]3 years ago

7 0

Answer:

Explanation:

Field by negative charge will be more as net force due to both the charges is in positive y - direction.

field by negative charge at origin

= 9 x 10⁹ x 34 x 10⁻⁶ / .18²

= 9.45 x 10⁶ N /C

field by positive charge at origin

= 9 x 10⁹ x Q / .33²

= 82644.63 x 10⁶ Q N /C

Net field

= 9.45 x 10⁶ - 82644.63 x 10⁶ Q

force on charge 8.1 x 10⁻⁶ C

( 9.45 x 10⁶ - 82644.63 x 10⁶ Q ) X 8.1 x 10⁻⁶ = 31

76.545 - 669421.5 Q = 31

669421.5 Q = 45.545

Q = 45.545 / 669421.5

= 68 X 10⁻⁶ C

= 68 μC

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Answer:

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2. $30.96^{o}$

3. $2.1 m/s^{2}$

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