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.

Answer 2

Instead of computing each force separately with actual numbers, this approach first derives a general ratio formula in terms of the original charges, and then substitutes the given values.

1. F/8: Using Coulomb's law, the initial force is \( F = k\frac{|q_1 q_2|}{r^2} \) with \( q_1 = 4q \) and \( q_2 = -2q \), so \( F = k\frac{|4q\times(-2q)|}{r^2} = k\frac{8q^2}{r^2} \). After contact, since the spheres are identical, the total charge \( q_1+q_2 = 4q+(-2q) = 2q \) splits equally, giving each sphere a new charge of \( q' = q \). The new force is \( F' = k\frac{q\times q}{r^2} = k\frac{q^2}{r^2} \). The general ratio is \( \frac{F'}{F} = \frac{q^2}{8q^2} = \frac{1}{8} \), so \( F' = \frac{F}{8} \). This matches exactly.
2. F/4: This would only be correct if the new force were four times smaller, which would require the new charge product to be a quarter of the original product; but the actual ratio here works out to one-eighth, not one-quarter, so this option does not match the calculation.
3. F: This would mean the force stayed unchanged, which would only happen if the charge product before and after contact were the same; here the charge magnitudes change substantially (from \( 4q \) and \( 2q \) to \( q \) and \( q \)), so the force cannot remain the same.
4. F/2: This would apply if the new charge product were exactly half the old one, but \( q\times q = q^2 \) is one-eighth of \( 8q^2 \), not one-half, so this option is also incorrect.

The general charge-redistribution rule confirms that equal conducting spheres brought into contact always end up sharing the algebraic sum of their charges equally, and applying this here gives a force exactly one-eighth of the original.

Therefore, the correct answer is F/8.