Question

A large insulated sphere of radius \(r\), charged with \(Q\) C is placed in contact with a small insulated uncharged sphere of radius \(r_{1}\) and is then separated. The charge on the smaller sphere will now be

• A. \(Q(r+r_{1})\)
• B. \(Q(r-r_{1})\)
• C. \(\frac{Q}{(r+r_{1})}\)
• D. \(\frac{Qr_{1}}{(r+r_{1})}\)

Hint:

After contact both spheres are at the same potential, and the total charge is conserved. Solve for the charge on the smaller sphere.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
When two conductors touch, charge flows until their potentials are equal. Total charge stays \(Q\).

Step 2: Equal Potential
\(\dfrac{kq_{1}}{r}=\dfrac{kq_{2}}{r_{1}}\), so \(q_{1}=\dfrac{q_{2}r}{r_{1}}\).

Step 3: Charge Conservation
\(q_{1}+q_{2}=Q\Rightarrow q_{2}\left(\dfrac{r}{r_{1}}+1\right)=Q\).

Step 4: Solve
\(q_{2}=\dfrac{Qr_{1}}{r+r_{1}}\).

Final Answer:
The small sphere ends up with charge Q r1 divided by r plus r1. \[ \boxed{\text{(D) }\dfrac{Qr_{1}}{r+r_{1}}} \]