Question:

Two metal spheres of radii $r_1$ and $r_2$ having charges $q_1$ and $q_2$ are brought in contact. Which statement is NOT correct?

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For conducting spheres in contact: always equate potentials, not charges.
  • Total charge is conserved
  • Both spheres attain same potential
  • Final potential is $\frac{1}{4\pi\epsilon_0}\frac{(q_1+q_2)}{(r_1+r_2)}$
  • Final potential is $\frac{1}{4\pi\epsilon_0}\frac{(q_1+q_2)(r_1+r_2)}{r_1 r_2}$
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The Correct Option is C

Solution and Explanation

When two conducting spheres are connected, charge redistributes until both reach the same potential.

Step 1: Condition of equilibrium
\[ V_1 = V_2 \Rightarrow \frac{q'_1}{r_1} = \frac{q'_2}{r_2} \]

Step 2: Total charge conservation
\[ q'_1 + q'_2 = q_1 + q_2 \]

Step 3: Final potential
Common potential: \[ V = \frac{1}{4\pi\epsilon_0}\frac{q_1+q_2}{r_1 + r_2 \text{ (incorrect assumption)}} \] This is WRONG because potential depends on individual radii, not sum. Correct derivation gives: \[ V = \frac{1}{4\pi\epsilon_0}\frac{q_1+q_2}{\frac{r_1 r_2}{r_1 + r_2}} \] So option (C) is incorrect. Final Answer: \[ \boxed{(C)} \]
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