Let rho (r) = frac{Q}{pi R^4} r be the volume | vedclass

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(D) To find the electric field at a distance $r_1$ from the center,we first calculate the charge $q$ enclosed within a sphere of radius $r_1$.The char

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$\frac{Q r_1^2}{4\pi \varepsilon_0 R^4}$

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(D) To find the electric field at a distance $r_1$ from the center,we first calculate the charge $q$ enclosed within a sphere of radius $r_1$.The charge element $dq$ in a spherical shell of radius $r$ and thickness $dr$ is given by $dq = \rho(r) \cdot 4\pi r^2 dr$.Substituting $\rho(r) = \frac{Q}{\pi R^4} r$,we get:$dq = \left( \frac{Q}{\pi R^4} r \right) \cdot 4\pi r^2 dr = \frac{4Q}{R^4} r^3 dr$.The total charge $q$ enclosed within radius $r_1$ is:$q = \int_0^{r_1} \frac{4Q}{R^4} r^3 dr = \frac{4Q}{R^4} \left[ \frac{r^4}{4} \right]_0^{r_1} = \frac{Q r_1^4}{R^4}$.Using Gauss's Law for a spherical surface of radius $r_1$:$E \cdot 4\pi r_1^2 = \frac{q}{\varepsilon_0}$.Substituting the value of $q$:$E \cdot 4\pi r_1^2 = \frac{Q r_1^4}{\varepsilon_0 R^4}$.Solving for $E$:$E = \frac{Q r_1^4}{4\pi \varepsilon_0 R^4 r_1^2} = \frac{Q r_1^2}{4\pi \varepsilon_0 R^4}$.

Let $\rho (r) = \frac{Q}{\pi R^4} r$ be the volume 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:

Similar Questions

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$A$ conducting sphere of radius $R = 20 \ cm$ is given a charge $Q = 16 \ \mu C$. What is the electric field $\overrightarrow{E}$ at the centre?

In a uniformly charged sphere of total charge $Q$ and radius $R$,the electric field $E$ is plotted as a function of distance $r$ from the centre of the sphere. The graph which would correspond to the above description is:

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