Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Let  P(r)=\frac{Q}{\pi R^{4}}r  be the charge density distribution for a solid sphere of radius R and total charge Q . For a point 'p' inside the sphere at distance r_{1} from the centre of the sphere, the magnitude of electric field is

  • Option 1)

    0

  • Option 2)

    \frac{Q}{4\pi \varepsilon _{0}r_{1}^{2}}\; \;

  • Option 3)

    \; \frac{Qr_{1}^{2}}{4\pi \varepsilon _{0}R^{4}}\; \; \;

  • Option 4)

    \; \frac{Qr_{1}^{2}}{3\pi \varepsilon _{0}R^{4}}\;

 

Answers (2)

best_answer

As we learnt in

Gauss's Law -

Total flux linked with a closed surface called Gaussian surface.

Formula:

\phi = \oint \vec{E}\cdot d\vec{s}=\frac{Q_{enc}}{\epsilon _{0}}

 

- wherein

No need to be a real physical surface.

Qenc - charge enclosed by closed surface.

 

 If the charge density

\rho =\frac{Q}{\pi R^4}r

From Gauss Theorem

E(4\pi {r_1}^2) = \frac{q}{\varepsilon_o }

E(4\pi {r_1}^2) = \frac{1}{\varepsilon_o } \int \rho\: dV

E \:4\pi {r_1}^2 = \frac{1}{\varepsilon_o } \int_{0}^{r_1} \frac{Qr}{ \pi R^4} 4 \pi r^2 dr

\therefore E = \frac{Q{r_1}^2}{4 \pi \varepsilon_o R^4}

 


Option 1)

0

Incorrect

Option 2)

\frac{Q}{4\pi \varepsilon _{0}r_{1}^{2}}\; \;

Incorrect

Option 3)

\; \frac{Qr_{1}^{2}}{4\pi \varepsilon _{0}R^{4}}\; \; \;

Correct

Option 4)

\; \frac{Qr_{1}^{2}}{3\pi \varepsilon _{0}R^{4}}\;

Incorrect

Posted by

prateek

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

BTech Admission 2025 Live: JEE Mains, SRMJEEE, MET registrations underway; MHT CET syllabus updates
anam-khan

Centre issues coaching guidelines, prohibits centres from using false claims like '100% selection'
anam-khan

BTech Admissions 2025 Live: JEE Mains, SRMJEE, VITEEE dates out; AEEE, MET registrations underway

Latest Question