| ⇦ |
| ⇨ |
A total charge of 5µC is distributed uniformly on the surface of the thin walled semispherical cup. If the electric field strength at the centre of the hemisphere is 9×10⁸NC⁻¹, then the radius of the cup is
{1/4πε₀=9×10⁹N-m²C⁻²}
Options
(a) 5mm
(b) 10mm
(c) 5cm
(d) 10cm
Correct Answer:
5mm
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - Energy bands in solids are a consequence of
- Pure Si at 500 K has equal number of electron (nₑ) and hole (nₕ) concentrations
- In a ionised gas,the mobile charge carriers are
- The force F acting on a particle of mass m is indicated by the force-time graph
- The temperature coefficient of a resistance wire is 0.00125 per degree.
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Energy bands in solids are a consequence of
- Pure Si at 500 K has equal number of electron (nₑ) and hole (nₕ) concentrations
- In a ionised gas,the mobile charge carriers are
- The force F acting on a particle of mass m is indicated by the force-time graph
- The temperature coefficient of a resistance wire is 0.00125 per degree.
Topics: Electrostatics (146)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends

Leave a Reply