Q.

A non-conducting right circular hollow cone of base radius R has a uniform surface charge density σ . A part ABP is cut and removed from the cone. The potential due to the remaining portion of the cone at the apex point ‘P’ is  xy  σR∈0, where x, y are positive integers and co-primes. Find the value of  y−x ?

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answer is 7.

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Detailed Solution

Open the remaining cone, to form a part of disc of angle θ.Potential due to circular non conducting disc,  VC=σr2ε0 (where r is radius of disc)So,  VC=(θ2π)(σ⋅ℓ2ε0)=σℓ2ε053  πR(ℓ)12π=5σR12ε0
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