Question

Two point charges \( q_1 \) and \( q_2 \) are placed at a distance \( r \) in vacuum. The force between them is \( F \). If the distance is doubled and both charges are halved, what will be the new force?

• A. \( \dfrac{F}{8} \)
• B. \( \dfrac{F}{4} \)
• C. \( \dfrac{F}{2} \)
• D. \( \dfrac{F}{16} \)

Hint:

Coulomb’s law is inversely proportional to the square of the distance and directly proportional to the product of charges. If charges are scaled by \( a \) and distance by \( b \), the force scales by \( \dfrac{a^2}{b^2} \).

Answer 1

The Correct Option is D

To solve this problem, we'll use Coulomb's Law, which states the force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by: 

\( F = k \cdot \frac{q_1 \cdot q_2}{r^2} \)

where \( k \) is Coulomb's constant. Initially:
\( F = k \cdot \frac{q_1 \cdot q_2}{r^2} \)
Now, the distance is doubled, and both charges are halved, so the new distance is \( 2r \) and the new charges are \( \frac{q_1}{2} \) and \( \frac{q_2}{2} \). The new force \( F' \) can be expressed as:

\( F' = k \cdot \frac{\left( \frac{q_1}{2} \right) \cdot \left( \frac{q_2}{2} \right)}{(2r)^2} \)
\( F' = k \cdot \frac{\frac{q_1 \cdot q_2}{4}}{4r^2} \)
\( F' = k \cdot \frac{q_1 \cdot q_2}{16r^2} \)

Thus, the new force \( F' \) compared to the original force \( F \) is:

\( F' = \frac{1}{16} \cdot k \cdot \frac{q_1 \cdot q_2}{r^2} \)
\( F' = \frac{1}{16} \cdot F \)

Therefore, the new force is \( \frac{F}{16} \).

Answer 2

Step 1: Use Coulomb's law. 
The electrostatic force between two charges is given by: \[ F = \dfrac{k q_1 q_2}{r^2} \] Step 2: Apply changes to the charges and distance. 
If both charges are halved: \( q_1' = \dfrac{q_1}{2}, \quad q_2' = \dfrac{q_2}{2} \) 
Distance is doubled: \( r' = 2r \) 
So the new force becomes: \[ F' = \dfrac{k \cdot \left(\dfrac{q_1}{2}\right) \cdot \left(\dfrac{q_2}{2}\right)}{(2r)^2} = \dfrac{k \cdot q_1 q_2}{4 \cdot 4 r^2} = \dfrac{1}{16} \cdot \dfrac{k q_1 q_2}{r^2} = \dfrac{F}{16} \]