Rankers Physics
Topic: Electrostatics
Subtopic: Coulomb's Law

Charge Q is divided into two parts which are then kept some distance apart. The force between them will be maximum if the two parts are :
Q/2 each
Q/4 and 3Q/4
Q/3 and 2Q/3
e and (Q – e), where e = electronic charge

Solution:

To maximize the force between two charges \( q_1 \) and \( q_2 \) obtained by dividing a charge \( Q \) into two parts, we can use Coulomb's law:

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

where \( r \) is the distance between them.

Steps:

1. Set up the variables:
Let \( q_1 = x \) and \( q_2 = Q - x \).

2. Express the force:
Substitute \( q_1 \) and \( q_2 \) into the formula:
\[
F = k \frac{x (Q - x)}{r^2}
\]

3. Maximize \( F \):
To find the maximum force, take the derivative of \( F \) with respect to \( x \) and set it to zero:
\[
\frac{dF}{dx} = k \frac{Q - 2x}{r^2} = 0
\]

Solving \( Q - 2x = 0 \) gives \( x = \frac{Q}{2} \).

4. Conclusion:
The force is maximum when each part is \( \frac{Q}{2} \).

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