Question

Consider the electric force between a pair of charged particles a certain distance apart. By Coulomb's Law, if the charge on one of the particles is tripled, the force is _____.

one-half as much
one-third as much
two times as much
three times as much

Answer 1

The electric force is three times as much

Explanation:

The magnitude of the electric force between two charged object is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, let's call F the initial force between the two charges:

$F=k\frac{q_1 q_2}{r^2}$

Later, the charge on one of the particles is tripled, so that

$q_1' = 3 q_1$

Therefore, we can calculate the new electric force between the particles:

$F'=k\frac{q_1' q_2}{r^2}=k\frac{(3q_1)q_2}{r^2}=3(k\frac{q_1 q_2}{r^2})=3F$

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Answer 2

Coulomb gave the inverse square law that tells about the quantity of force present between two objects who are not in motion and are charged electrically.

The force will be:

Option D. Three times as much

This force can be explained as:

The magnitude of the charged electric objects is given as:

$\rm F = k \dfrac {q_{1}q_{2}}{r^{2}}$

Where,

• Coulomb constant (k) = $\rm 8.99 \times 10^{9} Nm^{-2}C^{-2}$

• Charges = q₁ and q₂

• Separation between the charged objects = r

• Let the initial force present between the charges = F

When the charge is increased by three times then,

$\rm q_{1}' = 3q_{1}$

Now, calculating the new electric force (F'):

$\begin{aligned}\rm F' &= \rm k\dfrac{\rm q_{1}'q_{2}}{\rm r^{2}}\\\\&= \rm k\dfrac{(\rm 3q_{1})q_{2}}{r^{2}}\\\\&=3(\rm k\dfrac{q_{1}q_{2}}{r^{2}})\end{aligned}$

We know that $\rm F = k \dfrac {q_{1}q_{2}}{r^{2}}$

$\therefore \rm F' = 3F$

Therefore, the force gets tripled.

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