Question

In: Physics

A charged particle A exerts a force of 2.62

Expert Solution

Force exerted by charged particle A on charged particle B is

F_AB = k Q_A Q_B / r²

where k = 8.99 * 10⁹ is the Coulomb's force constant

r = 13.7 * 10^-3 m is the separation between two charged particles

F_AB= 2.62 * 10^-6 N

Thus,

Q_AQ_B = 2.62 * 10^-6 * (13.7)² * 10 ^-6 /// / 8.99 * 10⁹ Eqn (1)

Since the charged particle A exerts force on charged particle B to the right, A is positively charged. And since the charged particle B moves away from the charged particle A, there is a repulsive force between the which implies, charged particle B is positive too.

Now, force exerted by the charged particle B on charged particle A at a distance 17.4 mm is

F_BA= k Q_A Q_B / r², where r = 17.4 mm

Inserting the value of Q_A Q_B from eqn (1), we get

F_BA= { 8.99 * 10⁹ * [2.62 * 10^-6 * (13.7)² * 10 ^-6 / 8.99 * 10⁹] } / [(17.4)² *10^-6]

On simplifying,

F_BA= 1.624 * 10^-6 N = 1.624 N

Thus, force on charged particle A by the charged particle B is 1.624 N and the direction of the force is to the left, as charged particle B has a positive charge.

genius_generous answered 2 years ago

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