Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Consider a sphere of radius R which carries a uniform charge density . If a sphere of radius <img alt="\frac{R}{2}" src="https://learn.careers360.com/latex-image/?%5Cfrac%7

Consider a sphere of radius R which carries a uniform charge density \rho. If a sphere of radius \frac{R}{2} is carved out of it, as shown, the ratio \frac{\left |\hat{E_A}\right |}{\left|\hat{E_B}\right |} of magnitude of electric field \hat{E_A} and \hat{E_B}, respectively, at points A and B due to the remeaning portion is:
Option: 1 \frac{21}{34}
Option: 2 \frac{18}{54}
Option: 3 \frac{17}{54}
Option: 4 \frac{18}{34}
 

Answers (1)

best_answer

 

 

 

 

 

 

Condider a sphere of density -\rho and radius R/2 is removed from a sphere of radius R

 

then, Electric field at A is E_A = E_\rho + E_{-\rho} = 0 -\frac{\rho(\frac{R}{2})}{3\epsilon_0} = \frac{\rho R}{6\epsilon_0}

electric field at B is E_B = E_\rho + E_{-\rho} = \frac{\rho(R)}{3\epsilon_0} - \frac{k\rho\frac{4}{3}\pi\left(\frac{R}{2} \right )^3}{\left(\frac{3R}{2} \right )^2} = \frac{17\rho R}{54\epsilon_0}

    \left |\frac{E_A}{E_B} \right| = \frac{9}{17} \Rightarrow \frac{18}{34}

Hence the correct option is (4). 

Posted by

avinash.dongre

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE aspirant in Kota commits suicide; 16th suicide since January
anam-khan

JEE Main 2025 Registration Live: NTA session 1 application last date today; official website, syllabus
anam-khan

JEE Main 2025 session 1 registration ends today at jeemain.nta.nic.in; application correction from November 26

Latest Question