Let P( r ) = Q \over \pi R^4r be the change 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 center of the sphere, the magnitude of electric field is :

Qr_1^2 \over 4\pi \, \in _0\,R^4

Qr_1^2 \over 3\pi \, \in _0\,R^4

Electrostatics PYQ — JEE Main 2009 · Shift 0 · Q63 with Solution

View written solutionFree

Correct answer: B

Given charge density

The volume charge density is

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

for a solid sphere of radius $R$ .

We need the electric field at a point inside the sphere, at distance $r_1$ from the center.

Use Gauss's law

Because the charge density depends only on distance $r$ from the center, the distribution is spherically symmetric.

So for a Gaussian surface of radius $r_1$ ,

E(4\pi r_1^2)=\frac{Q_{\text{enc}}}{\varepsilon_0}

Hence,

E=\frac{Q_{\text{enc}}}{4\pi\varepsilon_0 r_1^2}

So we first find the enclosed charge $Q_{\text{enc}}$ inside radius $r_1$ .

Find enclosed charge

Take a thin spherical shell of radius $r$ and thickness $dr$ . Its volume is

dV=4\pi r^2\,dr

Charge in this shell:

dq=\rho(r)\,dV

dq=\left(\frac{Q}{\pi R^4}r\right)(4\pi r^2dr)

dq=\frac{4Q}{R^4}r^3dr

Therefore,

Q_{\text{enc}}=\int_0^{r_1} \frac{4Q}{R^4}r^3dr

Q_{\text{enc}}=\frac{4Q}{R^4}\left[\frac{r^4}{4}\right]_0^{r_1}

Q_{\text{enc}}=\frac{Qr_1^4}{R^4}

Substitute into Gauss's law

E=\frac{Q_{\text{enc}}}{4\pi\varepsilon_0 r_1^2}

E=\frac{\frac{Qr_1^4}{R^4}}{4\pi\varepsilon_0 r_1^2}

E=\frac{Qr_1^2}{4\pi\varepsilon_0 R^4}

Match with options

This matches:

\boxed{\frac{Qr_1^2}{4\pi\varepsilon_0 R^4}}

So the correct option is B.

Comparison with stored answer

Stored correct answer: B

My derived answer: B

They agree.

More from Electrostatics

View written solutionFree

Correct answer: B

Given charge density

The volume charge density is

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

for a solid sphere of radius $R$ .

We need the electric field at a point inside the sphere, at distance $r_1$ from the center.

Use Gauss's law

Because the charge density depends only on distance $r$ from the center, the distribution is spherically symmetric.

So for a Gaussian surface of radius $r_1$ ,

E(4\pi r_1^2)=\frac{Q_{\text{enc}}}{\varepsilon_0}

Hence,

E=\frac{Q_{\text{enc}}}{4\pi\varepsilon_0 r_1^2}

So we first find the enclosed charge $Q_{\text{enc}}$ inside radius $r_1$ .

Find enclosed charge

Take a thin spherical shell of radius $r$ and thickness $dr$ . Its volume is

dV=4\pi r^2\,dr

Charge in this shell:

dq=\rho(r)\,dV

dq=\left(\frac{Q}{\pi R^4}r\right)(4\pi r^2dr)

dq=\frac{4Q}{R^4}r^3dr

Therefore,

Q_{\text{enc}}=\int_0^{r_1} \frac{4Q}{R^4}r^3dr

Q_{\text{enc}}=\frac{4Q}{R^4}\left[\frac{r^4}{4}\right]_0^{r_1}

Q_{\text{enc}}=\frac{Qr_1^4}{R^4}

Substitute into Gauss's law

E=\frac{Q_{\text{enc}}}{4\pi\varepsilon_0 r_1^2}

E=\frac{\frac{Qr_1^4}{R^4}}{4\pi\varepsilon_0 r_1^2}

E=\frac{Qr_1^2}{4\pi\varepsilon_0 R^4}

Match with options

This matches:

\boxed{\frac{Qr_1^2}{4\pi\varepsilon_0 R^4}}

So the correct option is B.

Comparison with stored answer

Stored correct answer: B

My derived answer: B

They agree.

Electrostatics question

Electrostatics question

More from Electrostatics

Electrostatics question

Electrostatics question

More from Electrostatics