Figure shows a solid hemisphere with a charge of 5nC distributed uniformly throughout its volume. The hemisphere lies on a plane and point P is located on the plane, along a radial line from the centre of curvature at distance 15 cm. The electric potential(volts) at point P due to the hemisphere, is: $(\frac{1}{4 π ε_{0}} = 9 \times 10^{9} (S . I u n i t s))$

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

answer is 300.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

By argument of symmetry, it will be half of the potential produced by the full sphere

$∵$ Charge on hemisphere $= Q$ , so charge on sphere $= 2 Q$

$\Rightarrow V = \frac{1}{2} . \frac{K (2 Q)}{R} = \frac{K Q}{R}$

$V = \frac{K Q}{R} = \frac{9 \times 10^{9} \times 5 \times 10^{- 9}}{15 \times 10^{- 2}} = 300 V$

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

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

Take Free Test