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```                   a point charge q is placed at the center of an uncharged conducting shell of inner radius R and outer radius 2R. if the point charge is removed from the centre of the shell, to infinity work done by electric force
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2 years ago

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```                                        $\small V_{\infty}=0\\ V_{inside}=(-Q/4\pi \epsilonR +Q/4\pi \epsilon2R)=-Q/4\pi \epsilon2R$Because charge appearing on the inner surface will be -Q and since the charge was originally zero so charge on outside =+Q.$\small W=-\Delta V \\=>W=-(V_{\infty}-V_0)=V_{0}$Thanks & RegardsArun KumarIIT DelhiAskiitians Faculty
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one year ago

There is a cube of side a at every corners of which equal charge (say, q) is placed. Find the force acting on any one of the charges?

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 Neeti 4 months ago

is it 4*3 1/2 kq 2 /a 2 ???

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CALCULATE THAT.....!

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what is intensityb & what is tha average power

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the force acting on moving charge particle in a uniform magnetic field is (a) normal to magnetic field (b) at 45 degree from the magnetic field (c) parallel to the magnetic field

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