| ⇦ |
| ⇨ |
A hollow sphere of radius 0.1m has a charge of 5×10⁻⁸C. The potential at a distance of 5cm from the centre of the sphere is
(1/4πε₀=9×10⁹NM²C⁻²)
Options
(a) 4000V
(b) 4500V
(c) 5000V
(d) 6000V
Correct Answer:
4500V
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - The approxiamate ratio of nuclear mass densities of ₇₉Au¹⁹⁷ and ₄₇ Ag¹⁰⁷ nuclei is
- A spherical body of emissivity e=0.6, placed inside a perfectly black body is maintained
- According to photon theory of light which of the following physical quantities,
- The displacement of a particle varies according to the relation x=4 (cos π t + sin π t)
- The charges Q, +q, and +q are placed at the vartices of an equilateral
Topics: Electrostatics
(146)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The approxiamate ratio of nuclear mass densities of ₇₉Au¹⁹⁷ and ₄₇ Ag¹⁰⁷ nuclei is
- A spherical body of emissivity e=0.6, placed inside a perfectly black body is maintained
- According to photon theory of light which of the following physical quantities,
- The displacement of a particle varies according to the relation x=4 (cos π t + sin π t)
- The charges Q, +q, and +q are placed at the vartices of an equilateral
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

Given,R=0.1m,q=5×10^-8c
V=kq/R=>V=9×10^9×5×10^-8/0.1=>V=45×10/0.1=>V=450×10=>V=4500v