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3 years ago

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Two point charges +3 micro coulomb and +8 micro coulomb repel each other with a force of 40N . If a charge of -5 micro coulomb i

s added to each of them, then the force between them will become?

1 answer:

fgiga [73]3 years ago

5 0

Answer:

1 N

Explanation:

Given that two point charges +3 micro coulomb and +8 micro coulomb repel each other with a force of 40N.

Let us first calculate the distance between the two point charges

F = (KQq)/r^2

Where K = 9.0 × 10^9Nm^2C^-2

40 = (9×10^9 × 3×10^-6 × 8×10^-6)/r^2

Make r^2 the subject of formula

r^2 = 0.216/40

r = sqrt (5.4^-3)

r = 0.0735 metres

If a charge of -5 micro coulomb is added to each of them, then the force between them will become?

Using the same formula again

F = ( KQq)/r^2

Where Q = 8 - 5 = +3 × 10^-6 C

q = 3 - 5 = - 2 × 10^-6 C

F = ( 9×10^9 × 3×10^-6 × 2×10^-6 )/5.4×10^-2

F = 0.054/0.054

F = 1 N

The force of attraction between the two point charges will be 1 N

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$\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }$

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$\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}$

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$\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }$

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$\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}$

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