Question

If n electrons from a neutral solid sphere are transferred to another solid sphere having m electrons (m $>$ n), then the charges on the respective spheres are (e = charge of an electron)

• A. +ne and -(m-n)e
• B. -ne and +(m-n)e
• C. +ne and +(m-n)e
• D. +ne and -(m+n)e
• E. -ne and -(m+n)e

Hint:

Total charge is conserved.
Initial total charge = \( 0 + (-me) = -me \).
Final total charge = \( +ne + [-(m+n)e] = ne - me - ne = -me \).
This confirms the conservation of charge.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Charging by friction or transfer involves the movement of electrons.
A body that loses electrons becomes positively charged, while a body that gains electrons becomes negatively charged.

Step 3: Detailed Explanation:
1. Sphere 1: Initially neutral (Charge = 0). It loses \( n \) electrons.
According to the principle of quantization of charge, its new charge is \( q_1 = +ne \).
2. Sphere 2: Initially has \( m \) electrons. If we assume the sphere was neutral in terms of protons, its initial charge is \( q_i = -me \).
It gains \( n \) more electrons from the first sphere.
The final charge \( q_2 = q_i + (-ne) = -me - ne = -(m+n)e \).
Thus, the charges are \( +ne \) and \( -(m+n)e \).

Step 4: Final Answer:
The charges on the spheres are \( +ne \) and \( -(m+n)e \).