SATHEE: Electrostatics Question 342

Electrostatics Question 342

Question: Two spheres $A$ and $B$ of radius $4 c m$ and $6 c m$ are given charges of $80 μ c$ and $40 μ c$ respectively. If they are connected by a fine wire, the amount of charge flowing from one to the other is [MP PET 1991]

Options:

A) $20 μ C$ from $A$ to $B$

B) $16 μ C$ from $A$ to $B$

C) $32 μ C$ from $B$ to $A$

D) $32 μ C$ from $A$ to $B$

Show Answer

Answer:

Correct Answer: D

Solution:

Total charge $Q = 80 + 40 = 120 μ C$ .

By using the formula $Q_{1}^{'} = Q [\frac{r_{1}}{r_{1} + r_{2}}]$ . New charge on sphere A is $Q_{A}^{^{'}} = Q [\frac{r_{A}}{r_{A} + r_{B}}]$

$= 120 [\frac{4}{4 + 6}] = 48 μ C$ .

Initially it was $80 μ C$ i.e., $32 μ C$ charge flows from A to B.

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Electrostatics Question 342

Question: Two spheres A and B of radius 4cm and 6cm are given charges of 80μc and 40μc respectively. If they are connected by a fine wire, the amount of charge flowing from one to the other is [MP PET 1991]

Options:

Answer:

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Question: Two spheres $A$ and $B$ of radius $4 c m$ and $6 c m$ are given charges of $80 μ c$ and $40 μ c$ respectively. If they are connected by a fine wire, the amount of charge flowing from one to the other is [MP PET 1991]