| ⇦ | 
| ⇨ |
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: - If a magnet is suspended at an angle 30⁰ to the magnetic meridian, the dip needle
- Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube.
- When a mass M is attached to the spring of force constant k,
- The rms current in an AC circuit is 2 A. If the wattless current be √3 A,
- The angle between the dipole moment and electric field at any point on the equatoria
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- If a magnet is suspended at an angle 30⁰ to the magnetic meridian, the dip needle
- Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube.
- When a mass M is attached to the spring of force constant k,
- The rms current in an AC circuit is 2 A. If the wattless current be √3 A,
- The angle between the dipole moment and electric field at any point on the equatoria
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