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 $\frac{x}{y} \frac{σ R}{\in_{0}}$ , where x, y are positive integers and co-primes. Find the value of $y - x$ ?

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 7.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Open the remaining cone, to form a part of disc of angle $θ$ .

Potential due to circular non conducting disc, $V_{C} = \frac{σ r}{2 ε_{0}}$ (where r is radius of disc)
So, $V_{C} = (\frac{θ}{2 π}) (\frac{σ \cdot ℓ}{2 ε_{0}})$ $= (\frac{σ ℓ}{2 ε_{0}}) (\frac{5}{3} \frac{π R}{(ℓ)}) (\frac{1}{2 π}) = \frac{5 σ R}{12 ε_{0}}$

Watch 3-min video & get full concept clarity

tricks from toppers of Infinity Learn

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET