Question

Two identical conducting spheres, one carrying a charge of +4q and the other carrying -2q, are kept at a distance r apart in a vacuum. The magnitude of the electrostatic force between them is F. If they are brought into contact and then separated to the same distance r, what is the new electrostatic force between them?

• A. F/8
• B. F/4
• C. F
• D. F/2

Hint:

For problems involving charge redistribution on identical conducting spheres, remember that the total charge is conserved and distributes equally. The magnitude of force from Coulomb's law uses the absolute value of the product of charges.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem involves two charged conducting spheres. We need to calculate the initial electrostatic force, then determine the new charges on the spheres after they touch and separate, and finally calculate the new electrostatic force and its ratio to the initial force.

Step 2: Key Formula or Approach:
1. Coulomb's Law: The electrostatic force ($F$) between two point charges ($q_1, q_2$) separated by distance ($r$) is:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where $k = \frac{1}{4\pi\epsilon_0}$.
2. Charge Redistribution on Contact: When identical conducting spheres are brought into contact, charge redistributes evenly between them. The final charge on each sphere is the total charge divided by the number of spheres.

Step 3: Detailed Explanation:
Initial State:
- Charges: $q_1 = +4q$, $q_2 = -2q$.
- Distance: $r$.
- Initial Force ($F$):
\[ F = k \frac{|(+4q)(-2q)|}{r^2} = k \frac{|-8q^2|}{r^2} = k \frac{8q^2}{r^2} \]
After Contact and Separation:
- When the two identical conducting spheres are brought into contact, the total charge is distributed equally between them.
- Total charge = $q_1 + q_2 = +4q + (-2q) = +2q$.
- Since the spheres are identical, after separation, each sphere will have a charge of $\frac{+2q}{2} = +q$.
- New charges: $q'_1 = +q$, $q'_2 = +q$.
- Distance: $r$ (same as initial).
- New Force ($F'$):
\[ F' = k \frac{|(+q)(+q)|}{r^2} = k \frac{q^2}{r^2} \]
Ratio of Forces:
Now, let's find the ratio of the new force to the initial force:
\[ \frac{F'}{F} = \frac{k \frac{q^2}{r^2}}{k \frac{8q^2}{r^2}} = \frac{q^2}{8q^2} = \frac{1}{8} \]
So, $F' = \frac{F}{8}$.

Step 4: Final Answer:
The new electrostatic force between them is F/8.