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Question: A space is filled up with volume charge density \(\rho = {\rho _0}{e^{ - \alpha {r^3}}}\) where \({\...

A space is filled up with volume charge density ρ=ρ0eαr3\rho = {\rho _0}{e^{ - \alpha {r^3}}} where ρ0{\rho _0} and α\alpha are positive constants, r is the distance from the centre of the system. The magnitude of electric field strength as a function of r is given as ρ0xε0αr2(1eαr3)\dfrac{{{\rho _0}}}{{x{\varepsilon _0}\alpha {r^2}}}\left( {1 - {e^{ - \alpha {r^3}}}} \right) . Find x.

Explanation

Solution

We can solve this problem using gauss law. According to gauss law the flux through a closed surface is 1ε0\dfrac{1}{{{\varepsilon _0}}} times the charge enclosed by it. Using the gauss law equation, we can calculate the electric field. By comparing it with the equation for the electric field given in the question we can find the value of x.

Complete step by step answer:
It is given that a space is filled with volume charge density
ρ=ρ0eαr3\rho = {\rho _0}{e^{ - \alpha {r^3}}}
Where ρ0{\rho _0} and α\alpha are positive constants.
The distance from the centre of the system is given as r.
We need to calculate the magnitude of electric field strength as a function of r so that we can find the value of x.
Let us use gauss law to find the electric field.
we know that according to gauss law the flux through a closed surface will be equal to 1ε0\dfrac{1}{{{\varepsilon _0}}} times the total charge enclosed by the closed surface.
Here we can consider a spherical surface of radius r as the gaussian surface.
Gauss law can be written as
SEndA=1ε0Qinside\oint_S {E_n dA = \frac{1}{{\varepsilon _0 }}} Q_{inside} (1)
Where, E is the electric field and q is the total charge enclosed.
The volume charge density is the charge per unit volume
So, the total charge enclosed can be found by multiplying the volume charge density with volume.
q=ρdV\Rightarrow q = \int \rho dV
Where, ρ\rho is the volume charge density and V is the volume.

If we consider sphere with radius rr and thickness drdr, then the volume can be written as
dV=4πr2×drdV = 4\pi {r^2} \times dr
Where, 4πr24\pi {r^2} is the surface area of the sphere.
So, we get,
q=0rρ×4πr2dr\Rightarrow q = \int\limits_0^r {\rho \times 4\pi {r^2}dr}
q=0rρ0eαr3×4πr2dr\Rightarrow q = \int\limits_0^r {{\rho _0}{e^{ - \alpha {r^3}}} \times 4\pi {r^2}dr}
q=4πρ030r3r2eαr3dr\Rightarrow q = \dfrac{{4\pi {\rho _0}}}{3}\int\limits_0^r {3{r^2}{e^{ - \alpha {r^3}}}dr}
Let us find the value of the integral
I=0r3r2eαr3drI = \int\limits_0^r {3{r^2}{e^{ - \alpha {r^3}}}dr}
Let us put
αr3=t- \alpha {r^3} = t
On taking the derivative on both sides, we get
3r2dr=dtα\Rightarrow - 3{r^2}dr = \dfrac{{ - dt}}{\alpha }
When r=0r = 0 , t=0t = 0
When r=rr = r , t=αr3t = - \alpha {r^3}
We can write the integral in terms of t as
I=0αr3etdtαI = \int\limits_0^{ - \alpha {r^3}} {{e^t}\dfrac{{ - dt}}{\alpha }}
I=1α0αr3etdt\Rightarrow I = \dfrac{{ - 1}}{\alpha }\int\limits_0^{ - \alpha {r^3}} {{e^t}dt}
I=1α[et]0αr3\Rightarrow I = \dfrac{{ - 1}}{\alpha }\left[ {{e^t}} \right]_0^{ - \alpha {r^3}}
I=1α[eαr3e0]\Rightarrow I = \dfrac{{ - 1}}{\alpha }\left[ {{e^{ - \alpha {r^3} - {e^0}}}} \right]
I=1α[1eαr3]\Rightarrow I = \dfrac{1}{\alpha }\left[ {1 - {e^{ - \alpha {r^3}}}} \right]
On substituting the value of integral in equation for charge enclosed, we get
q=4πρ031α[1eαr3]\Rightarrow q = \dfrac{{4\pi {\rho _0}}}{3}\dfrac{1}{\alpha }\left[ {1 - {e^{ - \alpha {r^3}}}} \right]
Now let us substitute this value in equation 1.

E×4πr2=4πρ03αε0[1eαr3] \Rightarrow E \times 4\pi {r^2} = \dfrac{{4\pi {\rho _0}}}{{3\alpha {\varepsilon _0}}}\left[ {1 - {e^{ - \alpha {r^3}}}} \right]
E=4πρ03αε0×4πr2[1eαr3]\Rightarrow E = \dfrac{{4\pi {\rho _0}}}{{3\alpha {\varepsilon _0} \times 4\pi {r^2}}}\left[ {1 - {e^{ - \alpha {r^3}}}} \right]
E=ρ03ε0αr2(1eαr3)\Rightarrow E = \dfrac{{{\rho _0}}}{{3{\varepsilon _0}\alpha {r^2}}}\left( {1 - {e^{ - \alpha {r^3}}}} \right) (2)
In the question the value of electric field is given as
E=ρ0xε0αr2(1eαr3)E = \dfrac{{{\rho _0}}}{{x{\varepsilon _0}\alpha {r^2}}}\left( {1 - {e^{ - \alpha {r^3}}}} \right) (3)
On comparing equation 3 and 2 we can see that the value of x is 3.
x=3\therefore x = 3

Note: Remember that gauss is not applicable in all cases. It is applied in situations where there is some symmetry in the charge distribution. The gaussian surface that is considered should also have some symmetry like the charge distribution so the electric field is constant across the gaussian surface.