Question

Two charges $Q_1$ and $-Q_2$ are separated by a distance $r$. The charges attract each other with a force $F$. What is the new force between the charges if the distance is cut to one-fourth and the magnitude of each charge is doubled?

• A. 16 \(F\)
• B. 64 \(F\)
• C. 48 \(F\)
• D. \( \frac{1}{48}F \)

Hint:

When the charges are doubled and the distance is reduced by a factor of four, the force increases by a factor of 64 according to Coulomb's law.

Answer 1

The Correct Option is B

Step 1: Using Coulomb’s Law
The force between two point charges is given by Coulomb’s law: \[ F = k \frac{|Q_1 Q_2|}{r^2} \] Where:
\(F\) is the force,
\(k\) is Coulomb's constant,
\(Q_1\) and \(Q_2\) are the magnitudes of the charges,
\(r\) is the distance between the charges.

Step 2: Effect of Changing Distance and Charges
When the distance is cut to one-fourth, the new distance \(r' = \frac{r}{4}\).
When the charges are doubled, the new charges are \(2Q_1\) and \(2Q_2\).
The new force \(F'\) is: \[ F' = k \frac{|(2Q_1)(2Q_2)|}{\left(\frac{r}{4}\right)^2} = k \frac{4|Q_1 Q_2|}{\frac{r^2}{16}} = 64 \times k \frac{|Q_1 Q_2|}{r^2} \]
Step 3: Conclusion
Thus, the new force between the charges is \(64F\).