Question

Force between two charges \( +8\,\mu C \) and \( +2\,\mu C \) is \( 16\,N \). If the charges are brought into contact and then separated by the same distance, the force between them is

• A. \( 25\,N \)
• B. \( 20\,N \)
• C. \( 30\,N \)
• D. \( 15\,N \)
• E. \( 12\,N \)

Hint:

When identical conductors touch, charges redistribute equally. Always find new charges first, then apply Coulomb’s law again.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem involves Coulomb's law and the principle of conservation of charge. First, we use the initial information to find the relationship between the force and the charges. Then, we find the new charges after they are brought into contact (charge redistribution) and calculate the new force.
Step 2: Key Formula or Approach:
1. Coulomb's Law: The force F between two point charges \( q_1 \) and \( q_2 \) separated by a distance r is given by \( F = k \frac{|q_1 q_2|}{r^2} \), where k is Coulomb's constant.
2. Charge Redistribution: When two identical conducting spheres are brought into contact, the total charge is shared equally between them. The new charge on each sphere will be \( q' = \frac{q_1 + q_2}{2} \).
Step 3: Detailed Explanation:
Initial Situation:
The initial charges are \( q_1 = +8 \, \mu\text{C} \) and \( q_2 = +2 \, \mu\text{C} \).
The initial force between them is \( F_1 = 16 \) N.
Using Coulomb's law:
\[ F_1 = k \frac{q_1 q_2}{r^2} \implies 16 = k \frac{(8 \times 10^{-6})(2 \times 10^{-6})}{r^2} = k \frac{16 \times 10^{-12}}{r^2} \] From this, we find that \( \frac{k}{r^2} = \frac{16}{16 \times 10^{-12}} = 10^{12} \) N/C\(^2\).
After Contact:
The charges are brought into contact. The total charge is \( Q_{total} = q_1 + q_2 = 8 \, \mu\text{C} + 2 \, \mu\text{C} = 10 \, \mu\text{C} \).
This total charge is shared equally between the two identical spheres. The new charge on each sphere is:
\[ q' = \frac{Q_{total}}{2} = \frac{10 \, \mu\text{C}}{2} = 5 \, \mu\text{C} \] So, the new charges are \( q'_1 = 5 \, \mu\text{C} \) and \( q'_2 = 5 \, \mu\text{C} \).
Final Situation:
The charges are separated by the same distance r. The new force \( F_2 \) is:
\[ F_2 = k \frac{q'_1 q'_2}{r^2} = k \frac{(5 \times 10^{-6})(5 \times 10^{-6})}{r^2} = k \frac{25 \times 10^{-12}}{r^2} \] We can find \( F_2 \) by using the ratio method to avoid calculating \( k/r^2 \).
\[ \frac{F_2}{F_1} = \frac{k \frac{q'_1 q'_2}{r^2}}{k \frac{q_1 q_2}{r^2}} = \frac{q'_1 q'_2}{q_1 q_2} \] \[ \frac{F_2}{16} = \frac{(5 \times 10^{-6})(5 \times 10^{-6})}{(8 \times 10^{-6})(2 \times 10^{-6})} = \frac{25}{16} \] \[ F_2 = 16 \times \frac{25}{16} = 25 \text{ N} \] Step 4: Final Answer:
The new force between the charges is 25 N. This corresponds to option (A).