A solid ball of radius R has a charge density $\rho$ given by $\rho=\rho_0(1-\frac{r}{R})$ for $0< r< R$. The electric field outside the ball is :
Electric Field Calculation Using Gauss's Law
The problem asks for the electric field ($E$) outside a solid ball of radius $R$ where the charge density varies with the radial distance $r$ according to $\rho = \rho_0(1 - \frac{r}{R})$ for $0 < r < R$. We need to find the electric field for $r > R$. Due to the spherical symmetry of the charge distribution, the electric field will be radial and its magnitude will depend only on the distance $r$ from the center.
Applying Gauss's Law
We use Gauss's Law, which states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space ($\epsilon_0$).
$ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} $
Choose a spherical Gaussian surface with radius $r > R$. The electric field $\vec{E}$ is perpendicular to the surface element $d\vec{A}$ and has a constant magnitude on this surface.
The left side of Gauss's Law becomes:
$ \oint \vec{E} \cdot d\vec{A} = E \oint dA = E (4\pi r^2) $
Calculating Enclosed Charge ($Q_{enc}$)
Since the Gaussian surface has a radius $r > R$, it encloses the entire charge of the solid ball. We need to calculate the total charge ($Q_{total}$) within the ball.
The total charge is found by integrating the charge density $\rho$ over the volume ($V$) of the ball:
$ Q_{total} = \int_V \rho dV $
Using spherical coordinates, $dV = 4\pi r^2 dr$. The integration limits are from $0$ to $R$.
$ Q_{total} = \int_0^R \rho_0 \left(1 - \frac{r}{R}\right) (4\pi r^2 dr) $
Factor out constants:
$ Q_{total} = 4\pi \rho_0 \int_0^R \left(r^2 - \frac{r^3}{R}\right) dr $
Perform the integration:
$ Q_{total} = 4\pi \rho_0 \left[ \frac{r^3}{3} - \frac{r^4}{4R} \right]_0^R $
Evaluate the limits:
$ Q_{total} = 4\pi \rho_0 \left( \frac{R^3}{3} - \frac{R^4}{4R} \right) $
$ Q_{total} = 4\pi \rho_0 \left( \frac{R^3}{3} - \frac{R^3}{4} \right) $
$ Q_{total} = 4\pi \rho_0 \left( \frac{4R^3 - 3R^3}{12} \right) $
$ Q_{total} = 4\pi \rho_0 \left( \frac{R^3}{12} \right) $
$ Q_{total} = \frac{\pi \rho_0 R^3}{3} $
So, the enclosed charge for $r > R$ is $Q_{enc} = Q_{total} = \frac{\pi \rho_0 R^3}{3}$.
Determining the Electric Field
Now, substitute the results back into Gauss's Law:
$ E (4\pi r^2) = \frac{Q_{enc}}{\epsilon_0} $
$ E (4\pi r^2) = \frac{1}{\epsilon_0} \left( \frac{\pi \rho_0 R^3}{3} \right) $
Solve for $E$:
$ E = \frac{1}{4\pi r^2} \frac{\pi \rho_0 R^3}{3 \epsilon_0} $
Simplify the expression:
$ E = \frac{\rho_0 R^3}{12 \epsilon_0 r^2} $
This is the electric field outside the ball ($r > R$).
Similar Questions
Two blocks of masses $m$ and $M$, ($M > m$), are placed on a frictionless table as shown in figure. A massless spring with spring constant $k$ is attached with the lower block. If the system is slightly displaced and released, then
A. The time period of small oscillation of the two blocks is $T = 2\pi \sqrt{\frac{(m+M)}{k}}$
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Important Questions from Electricity and Magnetism
Two blocks of masses $m$ and $M$, ($M > m$), are placed on a frictionless table as shown in figure. A massless spring with spring constant $k$ is attached with the lower block. If the system is slightly displaced and released, then
A. The time period of small oscillation of the two blocks is $T = 2\pi \sqrt{\frac{(m+M)}{k}}$
B. The acceleration of the blocks is $a = \frac{kx}{M+m}$ (x = displacement of the blocks from the mean position)
C. The magnitude of the frictional force on the upper block is $\frac{\mu m|x|}{M+m}$
D. The maximum amplitude of the upper block, if it does not slip, is $\frac{\mu (M+m)g}{k}$
E. Maximum frictional force can be $\mu(M + m) g$.
Choose the correct answer from the options given below:
A wire of length $25 \ m$ and cross-sectional area $5 \ mm^2$ having resistivity of $2 \times 10^{-6} \ \Omega \ m$ is bent into a complete circle. The resistance between diametrically opposite points will be
(DROPPED)
A parallel plate capacitor is filled equally(half) with two dielectrics of dielectric constants $\varepsilon_1$ and $\varepsilon_2$, as shown in figures. The distance between the plates is $d$ and area of each plate is $A$. If capacitance in first configuration and second configuration are $C_1$ and $C_2$ respectively, then $\frac{C_1}{C_2}$ is :
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