JEE Main202229 Jul 2022Morning ShiftPhysicsElectrostaticsActual
A spherically symmetric charge distribution is considered with charge density varying as ρ r = ρ 0 3 4 − r R for r ≤ R Zero for r > R Where, r r < R is the distance from the centre O (as shown in figure). The electric field at point P will be :
Options
- Aρ 0 r 4 ε 0 3 4 - r R
- Bρ 0 r 3 ε 0 3 4 - r R
- Cρ 0 r 4 ε 0 1 - r R
- Dρ 0 r 5 ε 0 1 - r R
Correct answer
C. ρ 0 r 4 ε 0 1 - r R
Step-by-step solution
According to Gauss's law, the electric flux through a closed surface area is equal to the charge inside the surface divided by ε 0 . ⇒ ∮ E · → d s → = Q in ε 0 ⇒ E 4 π r 2 = ∫ 0 r ρ 0 3 4 - r R 4 π r 2 d r ε 0 ⇒ E 4 π r 2 = ρ o 4 π ε o 3 4 r 3 3 - r 4 4 R ⇒ E = ρ o r 4 ε o 1 - r R