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mamaluj [8]
3 years ago
7

Two point charges, A and B, are separated by a distance of 19.0 cm . The magnitude of the charge on A is twice that of the charg

e on B. If each charge exerts a force of magnitude 46.0 N on the other, find the magnitudes of the charges
Physics
1 answer:
Hatshy [7]3 years ago
6 0

Answer:

The charge on point A is q_a = 2.4 × 10^{-5} C

The charge on point B is q_b = 1.2 × 10^{-5} C

Explanation:

Given data

Distance (r) = 0.19 m

Magnitude of the charge on A is twice that of the charge on B  i.e.

q_A = 2 q_B

F = 46 N

We know that force between the charges is given by

F = \frac{k Q_A Q_B}{r^{2} }

65 = \frac{(9) (10^{9}) (2q_b^{2} ) )}{0.2^{2} }

q_b = 1.2 × 10^{-5} C

q_a = 2 × q_b

q_a = 2 × 1.2 × 10^{-5}

q_a = 2.4 × 10^{-5} C

Therefore the charge on point A is q_a = 2.4 × 10^{-5} C

The charge on point B is q_b = 1.2 × 10^{-5} C

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Answer:

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Read 2 more answers
Did I do these questions correctly?
SOVA2 [1]
Yes, they seem right to me.
4 0
3 years ago
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