Two metallic spheres of radii 1 cm and 3 cm are given charges of -1x10-2 C and 5x10-2 C, respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is:

  • Option 1)

    3 x 10-2 C

  • Option 2)

    4 x 10-2 C

  • Option 3)

    1 x 10-2 C

     

  • Option 4)

    2 x 10-2 C

 

Answers (1)

As we learnt in 

Common Potential -

V=frac{Total: charge}{Total: capacity}=frac{Q_{1}+Q_{2}}{C_{1}+C_{2}}

V=frac{C_{1}V_{1}+C_{2}V_{2}}{C_{1}+C_{2}}

-

 

 V=\frac{q_{1}+q_{2}}{c_{1}+c_{2}}

= \frac{-1\times 10^{-2}+5\times 10^{-2}}{4\pi \varepsilon R_{1}+4\pi \varepsilon R_{2}}

=\frac{4\times }{4\pi \varepsilon _{0}\left [ 1\times 10^{-2}+3\times 10^{-2} \right ]}

final Charge on the finger space

= 4\pi \varepsilon _{0}\times 3\times 3\times 10^{-2}\times \frac{4\times 10^{-2}}{4\pi \varepsilon _{0}\times 4\times 10^{-2}}

= 3\times 10^{-2} C 


Option 1)

3 x 10-2 C

Correct

Option 2)

4 x 10-2 C

Incorrect

Option 3)

1 x 10-2 C

 

Incorrect

Option 4)

2 x 10-2 C

Incorrect

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