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