Q is the charge on the surface of the metal sphere having radius R . The electric potential at R/2 Inside the sphere is....
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Answer:
∴
4πε
0
R
q
1
′
=
4πε
0
(2R)
q
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))
q
1
′
=
3
5
σ4πR
2
;q
2
′
=2q
1
′
=
3
10
(σ4πR
2
)
σ
1
=
4πR
2
q
1
′
=
3
5
σ,σ
2
=
4π(2R)
2
q
2
′
=
6
5
σ
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