The electrostatics potential inside a charged spherical is given by $ϕ = a r^{2} + b$ where $r$ is the difference between centre $a, b$ are constants. Then the charge density of the ball is
$- 6 a ε_{o} r$
$- 24 π a ε_{o}$
$- 6 a ε_{o}$
$- 24 π a ε_{o} r$

Question

The electrostatics potential inside a charged spherical is given by $ϕ = a r^{2} + b$ where $r$ is the difference between centre $a, b$ are constants. Then the charge density of the ball is
$- 6 a ε_{o} r$
$- 24 π a ε_{o}$
$- 6 a ε_{o}$
$- 24 π a ε_{o} r$

Toppr Admin · Accepted Answer

E-x2212-d-x3D5-dr- E-x2212-2arand E-14-x3C0-x3F5-oqr2-x2212-2arq-x2212-4-x3C0-x3F5-o-x22C5-2ar3Charge density- -x3C1-qVolume-x2212-4-x3C0-x3F5-o2ar343-x3C0-r3-x3C1-x2212-6a-x3F5-o

The electrostatics potential inside a charged spherical is given by ϕ=ar2+b where r is the difference between centre a,b are constants. Then the charge density of the ball is−6aεor−24πaεo−6aεo−24πaεor

E=−dϕdr; E=−2arand E=14πϵoqr2=−2arq=−4πϵo⋅2ar3Charge density, ρ=qVolume=−4πϵo2ar343πr3ρ=−6aϵo

The electrostatics potential inside a charged spherical is given by $ϕ = a r^{2} + b$ where $r$ is the difference between centre $a, b$ are constants. Then the charge density of the ball is
$- 6 a ε_{o} r$
$- 24 π a ε_{o}$
$- 6 a ε_{o}$
$- 24 π a ε_{o} r$

$E = - \frac{d ϕ}{d r}$ ; $E = - 2 a r$
and $E = \frac{1}{4 π ϵ_{o}} \frac{q}{r^{2}} = - 2 a r$
$q = - 4 π ϵ_{o} \cdot 2 a r^{3}$
Charge density, $ρ = \frac{q}{V o l u m e} = \frac{- 4 π ϵ_{o} 2 a r^{3}}{\frac{4}{3} π r^{3}}$
$ρ = - 6 a ϵ_{o}$