A Ball Of Radius R Carries A Positive Charge Whose Volume Density Depends Only On A Separation R From The Ball's Centre As Ρ=Ρ 0(1−r/R), Where Ρ0 Is A Constant. Assuming The Permittivities Of The Ball And The Environment To Be Equal To Unity, Find The Following:

Sol:

(A) The Magnitude Of The Electric Field Strength As A Function Of The Distance R Both Inside And Outside The Ball.

Let us assume:

The ball to be divided into an infinite number of concentric thin shells of thickness = dr.

A shell at a radial distance = r.

Volume of this shell = pdV = \( 4\pi r^{2}\rho dr\) \( \Rightarrow 4\pi r^{2}\rho_{0}(1-\frac{r}{R})dr\)

Net charge enclosed in the sphere of radius r

\( q = \int pdv = \int_{0}^{r} 4\pi \rho_{0}r^{2} (1 – \frac{r}{R}) dr = 4\pi \rho_{0} (\frac{r^{3}}{3} – \frac{r^{4}}{4R})\)

The electric field at radial distance r

\( E = \frac{q}{4\pi \varepsilon _{0}r^{2}}\)

=\( \frac{q1}{4\pi \varepsilon _{0}r^{2}} * 4\pi \rho_{0}(\frac{r^{3}}{3} – \frac{r^{4}}{4R})\)

Now for those points outside the sphere, we have to consider the total charge contained in the sphere of radius R.

Total charge Q =\( \int_{0}^{R}\rho _{0} (1 – \frac{r}{R}) 4\pi^{2}dr = \frac{\pi \rho _{0}R^{3}}{3}\)

Electric field, using Gauss’s laws, at a point which is at distance r(>R) from the centre:

\( \int E.ds = \frac{Q}{\varepsilon _{0}}\) \( \Rightarrow E = \frac{Q}{4\pi \varepsilon _{0}r^{2}}\) \( \Rightarrow \frac{\rho _{0}R^{3}}{12\varepsilon _{0}r^{2}}\)

(B) The Maximum Intensity E Max And The Corresponding Distance R M.

Again let us consider the expression for electric field within the sphere of radius R:

\( E = \frac{\rho_{0}r}{3\varepsilon _{0}} (1 – \frac{3r}{4R})\)

For maximum electric field:

\( \frac{dE}{dr} = 0\) \(\Rightarrow 1 – \frac{6r}{4R} = 0\) \(\Rightarrow r = \frac{2R}{3}\)

Now by substituting r = \frac{2R}{3}, value of maximum electric field,

\(E_{max} = \frac{\rho _{0}R}{9\varepsilon _{0} }\)

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