Question

2.The electrostatic potential inside a charged spherical ball is given by phi=ar2+b where r is the distance from the centre ,a and b are constants.Then ,the charge density inside the ball is

Answer 1

Step 1

Given,

$\varnothing =a{r}^2+b$ Where a and b are cont.

r = distance from center.

Step 2

we can solve this by using gauss's law = $E\cdot 4\pi r^2=\frac{q_{in}}{{\in }_0}$

Step 3

Solve,

given, $\varnothing =a{r}^2+b$

Electrical filed E = $\frac{d\varnothing }{dr}=-2ar$

now putting the value of E on gauss's law

$E\cdot 4\pi r^2=\frac{q_{in}}{{\in }_0}$

$(-2ar)\cdot 4\pi r^2=\frac{q_{in}}{{\in }_0}$

$-8\pi {\in }_{0}a{r}^3=q_{in}$

$\frac{d{q}_{in}}{dr}=-24\pi {\in }_{0}a{r}^2$

now,

$V=\frac{4}{3}\pi r^3$

$\frac{dV}{dr}=4\pi r^3$

$Changingdensity\eta =\frac{d{q}_{in}}{dr}x\frac{dr}{dV}$

$=-\frac{24\pi {\in }_{0}a{r}^2}{4\pi r^2}=-6{\in }_{0}a$

Hence, the charge inside the ball is

$-6{\in }_{0}a$