A spherical charged conductor has a surface charge density $σ$ . The electric field on its surface is E and electric potential of the conductor is V. Now the radius of the sphere is halved keeping the charge to be constant. The new values of electric field and potential would be

2E,2V

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E,V/2

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4E,4V

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2E,4V

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Solution

The correct option is B E,V/2
For a spherical charged conductor;
$E = \frac{σ}{ε_{0}}$
$V = \frac{σ}{ε_{0}} σ$

If $σ^{'} = \frac{σ}{2}$
$E^{'} = \frac{σ}{ε_{0}} = E$
$V^{'} = \frac{σ}{ε_{0}} σ^{'} = \frac{V}{2}$

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A spherical charged conductor has a surface charge density σ. The electric field on its surface is E and electric potential of the conductor is V. Now the radius of the sphere is halved keeping the charge to be constant. The new values of electric field and potential would be

The correct option is B E,V/2For a spherical charged conductor;E=σε0V=σε0σIf σ′=σ2E′=σε0=EV′=σε0σ′=V2

A spherical charged conductor has a surface charge density $σ$ . The electric field on its surface is E and electric potential of the conductor is V. Now the radius of the sphere is halved keeping the charge to be constant. The new values of electric field and potential would be

The correct option is B E,V/2
For a spherical charged conductor;
$E = \frac{σ}{ε_{0}}$
$V = \frac{σ}{ε_{0}} σ$

If $σ^{'} = \frac{σ}{2}$
$E^{'} = \frac{σ}{ε_{0}} = E$
$V^{'} = \frac{σ}{ε_{0}} σ^{'} = \frac{V}{2}$