Two isolated conducting spheres S₁ and S₂ of radius (2/3)R and (1/3)R have 12µC and -3 µC charges, respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on S₁ and S₂ are respectively:

Option 1 - <p>+4.5µC and -4.5µC</p>
Option 2 - <p>4.5µC on both</p>
Option 3 - <p>6µC and 3µC<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 4 - <p>3µC and 6µC<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
6 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
A
5 months ago
Correct Option - 3
Detailed Solution:

When the two conducting spheres are connected by a wire, they will reach the same electric potential, V.
The total charge Q_total = 12µC + (-3µC) = 9µC. This total charge will redistribute.
Let the final charges be q? and q? + q? = 9µC.
The potential of a sphere is V = kq/r.
V? = V?
k q? /R? = k q? /R?

...Read more

Thumbs Up IconUpvote Thumbs Down Icon

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K
Colleges
|
1.2K
Exams
|
6.9L
Reviews
|
1.8M
Answers

Learn more about...

Physics Electrostatic Potential and Capacitance 2025

Physics Electrostatic Potential and Capacitance 2025

View Exam Details

Most viewed information

Summary

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

or

Ask Current Students, Alumni & our Experts

Have a question related to your career & education?

or

See what others like you are asking & answering