Two concentric sphere kept in air have radii R & r .
they have similar change and equal charge density σ. The electric potential at
their common centre is? ( MHT - CET ~ 2014 )
a
] σ
( R + r ) / ε o b ] σ ( R – r ) / ε o
c
] σ ( R + r ) / 2 ε o d ] σ ( R + r ) / 4 ε o
Solution:
Electric potential ( Definition ) : The
electric potential at a point in electric field is defined as the amount of
work done to bring unit positive charge from infinite distance to that point
against the direction of electric intensity.
I . e .
V = w / q o
Where q o is a unit
charge -
- - - - - - - - ( 1 )
But ,
W = q / ( 4 π ε o
r )
Putting this value in above equation we
get,
V = q / ( 4 π ε o
r ) -
- - - - - - - - ( 2 )
The potential at center due to sphere 1
is
V1 = q / ( 4 π ε o R
) - - - - - -
- - - ( 3 )
And due to sphere 2 is:
V2 = q / ( 4 π ε o r
) -
- - - - - - - - ( 4 )
Therefore the electric potential at
common centre can be written as
V = V 1 + V 2
V = [ q / ( 4 π ε o R
) ] + [ q / ( 4 π ε
o r ) ]
V = [ q / ( 4 π ε o) ] [ ( 1 / R ) + ( 1 /
r ) ]
V = [ q
/ ( 4 π ε o)
] [ ( R + r ) / ( R r ) ] - - - - - - - - - ( 5 )
But σ is surface charge density i . e .
σ = ( total charge on the surface
of sphere ‘ q ‘ ) / ( surface area of sphere )
Therefore ,
σ = Q / A
σ = q
/ ( 4 π R )
Equation ( 5 ) becomes
V = σ ( R +
r ) / ε o
The correct option is : option [ a ]
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