A uniformly charged hemisphere of radius b and charge density ρ has a hemispherical cavity of radius a a = b 2 cut from its centre. If the potential at the centre of the cavity is n ρ b 2 16 ∈ 0 then n = ?

V = ∫ a b K 2 π x 2 d x ρ x V = 2 π k ρ b 2 - a 2 2 = π ρ k b 2 - a 2 b = 2 a v = 3 π k ρ b 2

Start Free Mock LoginAlready registered? Login here

Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsPotential & potential energyEasy2 minPYQ_2020

PhysicsEasysingle choice

A uniformly charged hemisphere of radius $b$ and charge density $ρ$ has a hemispherical cavity of radius $a (a = \frac{b}{2})$ cut from its centre. If the potential at the centre of the cavity is $\frac{n ρ b^{2}}{16 \in_{0}}$ then $n = ?$

Options:

Answer:

Solution:

$V = \int_{a}^{b} \frac{K (2 π x^{2} d x) ρ}{x}$

$V = 2 π k ρ [\frac{b^{2} - a^{2}}{2}] = π ρ k (b^{2} - a^{2})$

$b = 2 a$

$v = 3 π k ρ b^{2}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Potential & potential energy

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...