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Ganezh [65]
3 years ago
9

Two small, positively charged spheres have a combined charge of 5.0 x 10 -5 C. If each sphere is repelled from the other by an e

lectrostatic force of 1.0 N when the spheres are 2.0 m apart, what is the charge, in micro-coulomb, on the sphere with the smaller charge?
Physics
1 answer:
geniusboy [140]3 years ago
4 0

Explanation:

It is given that,

Let q₁ and q₂ are two small positively charged spheres such that,

q_1+q_2=5\times 10^{-5}\ C.............(1)

Force of repulsion between the spheres, F = 1 N

Distance between spheres, d = 2 m

We need to find the charge on the sphere with the smaller charge. The force is given by :

F=k\dfrac{q_1q_2}{d^2}

q_1q_2=\dfrac{F.d^2}{k}

q_1q_2=\dfrac{1\ N\times (2\ m)^2}{9\times 10^9}

q_1q_2=4.45\times 10^{-10}\ C............(2)

On solving the system of equation (1) and (2) using graph we get,

q_1=0.0000384\ C=38.4\ \mu C

q_2=0.0000116\ C=11.6\ \mu C

So, the charge on the smaller sphere is 11.6 micro coulombs. Hence, this is the required solution.

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Now let's come to the Gravitational Force, we know that gravitational force is directly proportional to the mass of the body and inversely proportional to the radius.

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