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Fiesta28 [93]
3 years ago
10

Each of two small non-conducting spheres is charged positively, the combined charge being 40 μC. When the two spheres are 50 cm

apart, each sphere is repelled from the other by a force of magnitude 2.0 N. Determine the magnitude of the smaller of the two charges.
Physics
1 answer:
Phantasy [73]3 years ago
7 0

Answer:

1.44 x 10⁻⁶ C

Explanation:

q_{1} = charge on one sphere

q_{2} = charge on other sphere

q = Total charge on the two spheres = 40 μC

q_{1}+ q_{2} = q

q_{1}+ q_{2} = 40 x 10⁻⁶

q_{1} = (40 x 10⁻⁶) - q_{2}                                   eq-1

r = distance between the two spheres = 50 cm = 0.50 m

F = magnitude of force between the two spheres = 2.0 N

Magnitude of force between the two spheres is given as

F = \frac{k q_{1} q_{2}}{r^{2}}

2.0 = \frac{(9\times 10^{9}) ((40\times 10^{-6}) - q_{2}) q_{2}}{0.50^{2}}

q_{2} = 1.44 x 10⁻⁶ C

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brainly.com/question/3183125

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