Question

Two charges of Inc and
100nC are placed at a
distance of 10m in air.
Force between them is
9 x 10-9N
9 x 10°N
100N
10N​

Answer 1

Answer:

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Answer 2

Answer:

The force between them is 9×10⁻⁹N.

Explanation:

The force between two charges is calculated by Columbus's force. Which states that "The force of attraction or repulsion is directly proportional to the magnitude of the charges and inversely proportional to the square of the distance between the two charges". The force acts along the line joining the charges. Which can be given as,

$F_e\propto\frac{q_1 q_2}{d^2}$

$F_e=\frac{kq_1q_2}{d^2}$          (1)

Where,

K=columbs constant=9×10⁹Nm²C⁻²

q₁,q₂=charge on the particles

d= distance between the particles

The values given in the question are,

q₁=1nC=1×10⁻⁹C

q₂=100nC=100×10⁻⁹C

d=10m

By substituting the values in equation (1) we get;

$F_e=\frac{9\times 10^9\times 1\times 10^-^9\times 100\times 10^-^9}{(10)^2}$

$F_e=9\times 10^-^9N$

Hence, the force between the charges 1nC and 100nC is 9×10⁻⁹N.