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Grade: 12th pass

                        

Please sir reply to this question.Two metal spheres each of radius r and seperated by distance d,have equal and opposite charges.If d>>r the capacitance of the system is(I have doubt C=Q/V,but here qnet=0 as two equal and oopposite charges are present.so the capacitance must be zero.but this is incorrect. why??)

one year ago

Answers : (2)

Arun
24736 Points
							
Before contact, charges of each spheres,
1
​ 
 =σ4πR 
2
 ;q 
2
​ 
 =σ4π(2R) 
2
 =4q 
1
​ 
  When the two sphere are brought in contact, their charges are shared till their potentials become equal i.e., V 
1
​ 
 =V 
2
​ 
 
 
∴ 
4πε 
0
​ 
 R
1
​ 
 
​ 
 = 
4πε 
0
​ 
 (2R)
2
​ 
 
​ 
 
 
∴q 
2
​ 
 =2q 
1
​ 
 ....(i)
As there is no loss of charge in the process
 
∴q 
1
​ 
 +q 
2
​ 
 =q 
1
​ 
 +q 
2
​ 
 =q 
1
​ 
 +4q 
1
​ 
 =5q 
1
​ 
 =5(σ4πR 
2
 )
 
or q 
1
​ 
 +2q 
1
​ 
 =5σ4πR 
2
  (using (i))
 
1
​ 
 = 
3
5
​ 
 σ4πR 
2
 ;q 
2
​ 
 =2q 
1
​ 
 = 
3
10
​ 
 (σ4πR 
2
 )
 
σ 
1
​ 
 = 
4πR 
2
 
1
​ 
 
​ 
 = 
3
5
​ 
 σ,σ 
2
​ 
 = 
4π(2R) 
2
 
2
​ 
 
​ 
 = 
6
5
​ 
 σBefore contact, charges of each spheres,
1
​ 
 =σ4πR 
2
 ;q 
2
​ 
 =σ4π(2R) 
2
 =4q 
1
​ 
  When the two sphere are brought in contact, their charges are shared till their potentials become equal i.e., V 
1
​ 
 =V 
2
​ 
 
 
∴ 
4πε 
0
​ 
 R
1
​ 
 
​ 
 = 
4πε 
0
​ 
 (2R)
2
​ 
 
​ 
 
 
∴q 
2
​ 
 =2q 
1
​ 
 ....(i)
As there is no loss of charge in the process
 
∴q 
1
​ 
 +q 
2
​ 
 =q 
1
​ 
 +q 
2
​ 
 =q 
1
​ 
 +4q 
1
​ 
 =5q 
1
​ 
 =5(σ4πR 
2
 )
 
or q 
1
​ 
 +2q 
1
​ 
 =5σ4πR 
2
  (using (i))
 
1
​ 
 = 
3
5
​ 
 σ4πR 
2
 ;q 
2
​ 
 =2q 
1
​ 
 = 
3
10
​ 
 (σ4πR 
2
 )
 
σ 
1
​ 
 = 
4πR 
2
 
1
​ 
 
​ 
 = 
3
5
​ 
 σ,σ 
2
​ 
 = 
4π(2R) 
2
 
2
​ 
 
​ 
 = 
6
5
​ 
 σ
3 months ago
Vikas TU
12285 Points
							
Dear student 
The above explanation is not properly explained 
Please ignore this 
We are sorry for inconvenience.
Good Luck 
Cheers 
3 months ago
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