Question

A ball of radius R carries a positive charge whose volume density depends on separation r from centre of ball as ρ=k{1-r^2/R^2} where k is constant assuming permittivity of pollen environment to be equal to unity the intensity of the electric field is maximum at

Answer 1

Answer:

electric field is maximum at r = 0.745 R

Explanation:

As we know by Gauss law that electric flux due to a system of charge depends on the enclosed charge

So we have

$\int E. dA = \frac{q_{en}}{\epsilon_0}$

now we have

$E(4\pi r^2) = \frac{1}{\epsilon_0} \int k(1 - \frac{r^2}{R^2}) 4\pi r^2 dr$

$E r^2 = \frac{k}{\epsilon_0}(\frac{r^3}{3} - \frac{r^5}{5R^2})$

so we have

$E = \frac{k}{\epsilon_0}(\frac{r}{3} - \frac{r^3}{5R^2})$

now for maximum value of E we have

$\frac{dE}{dr} = 0$

so we have

$\frac{1}{3} - \frac{3r^2}{5R^2} = 0$

$r = 0.745 R$

So electric field is maximum at r = 0.745 R

#Learn

Topic : Electric field intensity

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