Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A point charge causes an electric flux of –1.0 × 10^3 Nm^2 /C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge.If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?

Q 1.20(a): A point charge causes an electric flux of -1.0\times 10^{3}\frac{Nm^{2}}{C}to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?

Answers (1)

best_answer

Given,

\phi = -1.0\times 10^{3}\frac{Nm^{2}}{C}

Using Gauss’s law, we know that the flux of electric field through any closed surface S is 1/\epsilon _{0}  times the total charge enclosed by S.

i.e.   \phi = q/\epsilon _{0}

where, q = net charge enclosed  and  \epsilon _{0} = permittivity of free space (constant)

Therefore, Flux does not depend on the radius of the sphere but only on the net charge enclosed. Hence, the flux remains the same although radius is doubled.

\phi' = -10^{3}\frac{Nm^{2}}{C}

Posted by

HARSH KANKARIA

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question