A thin sphericalal conducting shell of radius r has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point p at a distance r / 2 from the centre of the shell is

Question

Arun Kumar · Answer

Hi Ashu,

We will do this with gauss law.

for just a conducting shell field inside is zero since enclosed charge is 0

but here enclosed charge is Q

so

$\\E*4\pi \epsilon_{0} (r/2)^2=Q \\E=Q/4\pi \epsilon_{0} (r/2)^2$

Thanks & Regards

Arun Kumar

IIT Delhi

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A thin sphericalal conducting shell of radius r has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point p at a distance r / 2 from the centre of the shell is

A thin sphericalal conducting shell of radius r has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point p at a distance r / 2 from the centre of the shell is

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A thin sphericalal conducting shell of radius r has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point p at a distance r / 2 from the centre of the shell is

A thin sphericalal conducting shell of radius r has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point p at a distance r / 2 from the centre of the shell is

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