Charge is distributed within a sphere of radius R with a volume charge density - r - A - r 2 e - 2 r - a where A and a are constants- If Q is the total charge of this charge distribution- the radius R is -

According to the question Q=∫^R_0p(r)dv #160; #160; #160; #160; #160; [dv=4π r^2dr] ⇒ Q=∫^R_0Ar^2e^ 2r/a4π r^2dr Q=4π A∫^R_0e^ ...

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Standard XII

Physics

Area and Volume Expansion

Charge is dis...

Question

Charge is distributed within a sphere of radius R with a volume charge density $ρ (r) = \frac{A}{r^{2}} e^{- 2 r / a}$ where A and a are constants. If Q is the total charge of this charge distribution, the radius R is :

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Solution

According to the question
$Q = \int_{0}^{R} p (r) d v$ $[d v = 4 π r^{2} d r]$
$\Rightarrow Q = \int_{0}^{R} \frac{A}{r^{2}} e^{- 2 r / a} 4 π r^{2} d r$
$Q = 4 π A \int_{0}^{R} e^{- 2 r / a} d r = 4 A π {[- e^{- 2 r / a} \times \frac{a}{2}]}_{0}^{- 2 r / a} = \frac{4 π A a}{2} (- e^{- 2 R / a} + 1)$
$Q = 2 π A a (1 - e^{- 2 R / a})$
Find radius of the sphere
$1 - e^{- 2 R / a} = \frac{Q}{2 π A a}$
$1 - \frac{Q}{2 π A a} = e^{- 2 R / a}$
$l n (1 - \frac{Q}{2 π A a}) = \frac{- 2 R}{a}$
$\frac{- a}{2} l n (1 - \frac{Q}{2 π A a}) = R$
$R = \frac{a}{2} l n (\frac{2 π A a}{2 π A a - Q})$ .

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Charge is distributed within a sphere of radius R with a volume charge density ρ(r)=Ar2e−2r/a where A and a are constants. If Q is the total charge of this charge distribution, the radius R is :

According to the questionQ=∫R0p(r)dv [dv=4πr2dr]⇒Q=∫R0Ar2e−2r/a4πr2drQ=4πA∫R0e−2r/adr=4Aπ[−e−2r/a×a2]R0=4πAa2(−e−2R/a+1)Q=2πAa(1−e−2R/a)Find radius of the sphere1−e−2R/a=Q2πAa1−Q2πAa=e−2R/aln(1−Q2πAa)=−2Ra−a2ln(1−Q2πAa)=RR=a2ln(2πAa2πAa−Q).

Charge is distributed within a sphere of radius R with a volume charge density $ρ (r) = \frac{A}{r^{2}} e^{- 2 r / a}$ where A and a are constants. If Q is the total charge of this charge distribution, the radius R is :