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# 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??)

Arun
25763 Points
one year ago
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
Vikas TU
14149 Points
one year ago
Dear student
The above explanation is not properly explained