Question
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=ρ0e−αr3 where ρ0 and α 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 xε0αr2ρ0(1−e−αr3) . Find x.
Solution
We can solve this problem using gauss law. According to gauss law the flux through a closed surface is ε01 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=ρ0e−αr3
Where ρ0 and α 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 ε01 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=ε01Qinside (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
Where, ρ is the volume charge density and V is the volume.
If we consider sphere with radius r and thickness dr, then the volume can be written as
dV=4πr2×dr
Where, 4πr2 is the surface area of the sphere.
So, we get,
⇒q=0∫rρ×4πr2dr
⇒q=0∫rρ0e−αr3×4πr2dr
⇒q=34πρ00∫r3r2e−αr3dr
Let us find the value of the integral
I=0∫r3r2e−αr3dr
Let us put
−αr3=t
On taking the derivative on both sides, we get
⇒−3r2dr=α−dt
When r=0 , t=0
When r=r , t=−αr3
We can write the integral in terms of t as
I=0∫−αr3etα−dt
⇒I=α−10∫−αr3etdt
⇒I=α−1[et]0−αr3
⇒I=α−1[e−αr3−e0]
⇒I=α1[1−e−αr3]
On substituting the value of integral in equation for charge enclosed, we get
⇒q=34πρ0α1[1−e−αr3]
Now let us substitute this value in equation 1.
⇒E×4πr2=3αε04πρ0[1−e−αr3]
⇒E=3αε0×4πr24πρ0[1−e−αr3]
⇒E=3ε0αr2ρ0(1−e−αr3) (2)
In the question the value of electric field is given as
E=xε0αr2ρ0(1−e−αr3) (3)
On comparing equation 3 and 2 we can see that the value of x is 3.
∴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.