Let P( r )=(Q/π R4) 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 r1 from the centre of the sphere, the magnitude of the electric field is

Question

Let P( r )=(Q/π R4) 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 r1 from the centre of the sphere, the magnitude of the electric field is (A) Zero (B) (Q/4 π ϵ 0 r12) (C) (Q r12/4 π ϵ 0 R4) (D) (Q r12/3 π ϵ 0 R4)

Tardigrade Admin · Accepted Answer

P(.r.)=(Q/π R4)r <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-dqjhlhmic2lhe95v.jpg" /> oint E ⋅ ds =( q en /ε0)=(∫ ρ VdV /ε0) =( Qr 4 π r 2 dr /π R 4) / ε0 E. 4 π r12=((Q/π R4) 4 π (r14/4)/ε0) E =( Qr 12/4 π ε0 R 4)

Q. Let $P (r) = \frac{Q}{π R ^{4}} r$ 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 the electric field is

Zero

$\frac{Q}{4 π ϵ _{0} r _{1}^{2}}$

$\frac{Q r _{1}^{2}}{4 π ϵ _{0} R ^{4}}$

$\frac{Q r _{1}^{2}}{3 π ϵ _{0} R ^{4}}$

$P (r) = \frac{Q}{π R ^{4}} r$

$\oint E \cdot d s = \frac{q _{e n}}{ε _{0}} = \frac{\int ρ V d V}{ε _{0}}$
$= \frac{Q r 4 π r ^{2} d r}{π R ^{4}} / ε_{0}$
E. $4 π r_{1}^{2} = \frac{\frac{Q}{π R ^{4}} 4 π \frac{r _{1}^{4}}{4}}{ε _{0}}$
$E = \frac{Q r _{1}^{2}}{4 π ε _{0} R ^{4}}$

Q. Let P(r)=πR4Q​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 r1​ from the centre of the sphere, the magnitude of the electric field is

Zero

4πϵ0​r12​Q​

4πϵ0​R4Qr12​​

3πϵ0​R4Qr12​​

P(r)=πR4Q​r ∮E⋅ds=ε0​qen​​=ε0​∫ρVdV​ =πR4Qr4πr2dr​/ε0​ E. 4πr12​=ε0​πR4Q​4π4r14​​​ E=4πε0​R4Qr12​​

Q. Let $P (r) = \frac{Q}{π R ^{4}} r$ 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 the electric field is

$\frac{Q}{4 π ϵ _{0} r _{1}^{2}}$

$\frac{Q r _{1}^{2}}{4 π ϵ _{0} R ^{4}}$

$\frac{Q r _{1}^{2}}{3 π ϵ _{0} R ^{4}}$

$P (r) = \frac{Q}{π R ^{4}} r$

$\oint E \cdot d s = \frac{q _{e n}}{ε _{0}} = \frac{\int ρ V d V}{ε _{0}}$
$= \frac{Q r 4 π r ^{2} d r}{π R ^{4}} / ε_{0}$
E. $4 π r_{1}^{2} = \frac{\frac{Q}{π R ^{4}} 4 π \frac{r _{1}^{4}}{4}}{ε _{0}}$
$E = \frac{Q r _{1}^{2}}{4 π ε _{0} R ^{4}}$