What is magnitude of electric field?

Let rho(r) = (Q*r)/(pi*R^4) be the charge density distribution for a solid sphere of radius R and total charge Q.
for a point ā€˜pā€™ inside the sphere at distance r1 from the centre of the sphere, Find the magnitude of electric field.
Note : rho is Volume charge density
Please explain elaborately. Thanks in advance :)

2 Answers
Jul 6, 2016

|vec(E)| = (Q(r1)^2)/(4piepsilon_0R^4)

Explanation:

Ok we're going to use Gauss' law here. Imagine a Gaussian surface at radius r1 inside the sphere. The net flux leaving the surface is only affected by the parts of the sphere inside it.

oint vec(E)*dvec(A) = (Q_(enc))/(epsilon_(0))

By symmetry we can see that the electric field vector will always point radially outwards from the centre of the sphere which, incidentally, is the same direction as the area element at all points on our Gaussian surface. We can then write

vec(E)*dvec(A) = |vec(E)||dvec(A)|costheta

but theta = 0 so vec(E)*dvec(A) = |vec(E)||dvec(A)|

We now have:

|vec(E)| oint |dvec(A)| = (Q_(enc))/(epsilon_0)

ie |vec(E)||vec(A)| = (Q_(enc))/(epsilon_0)

We now need to work out the charge enclosed by our Gaussian surface. We have a volume charge density, so first we consider a spherical shell of thickness dr. This shell will have charge dQ given by

dQ = rho4pir^2dr

dQ = (ar)*4pir^2dr where a = Q/(piR^4)

Q_(enc) = 4api int_0^(r1) r^3 dr

Q_(enc) = (Q)/(piR^4)pi(r1)^4 = Q((r1)/(R))^4

implies |vec(E)||vec(A)| = (Q)/(epsilon_0)((r1)/(R))^4

implies |vec(E)| = (Q)/(epsilon_0*|vec(A)|)((r1)/(R))^4

|vec(E)| = (Q)/(epsilon_0*4pi(r1)^2)((r1)/(R))^4

|vec(E)| = (Q(r1)^2)/(4piepsilon_0R^4)

I think this makes sense, as at r1 = R it provides the expected electric field outside a sphere. I will admit my EM is a bit rusty though.

Dec 8, 2016

We use Gauss' law here.
Given is volume charge density rho=Q/(piR^4)r
We observe that by symmetry the electric field vector will always point radially outwards from the center of charged distribution.

Imagine a spherical Gaussian surface of a of radius r_1 inside charged distribution where point p is located. Where r_1 can have values from 0 " to "R.

We know that net flux leaving the Gaussian surface is contributed by the total charge inside this sphere of radius r_1.
Q_"eff" charge enclosed within sphere of radius of r_1 can be found by taking integral from limit r=0 " to " r_1

Let us we consider charge enclosed in infinitesimal volume dv having charge dQ in a spherical shell of thickness dr at radius r
dQ=rho4pir^2dr
=>dQ=Q/(piR^4)rxx4pir^2dr
=>dQ=Q/(piR^4)rxx4pir^2dr
=>dQ=(4Q)/(R^4)r^3dr

Therefore,
Q"eff"=int_0^(r_1)(4Q)/(R^4)r^3dr
=>Q"eff"=|(4Q)/(R^4)r^4/4|_0^(r_1)
=>Q"eff"=Q/(R^4)r_1^4

As the angle of electric field vector with any element area at the surface of sphere is 0^@, in the dot product we have costheta=1.

Therefore, from Gauss's law we obtain the electric flux Phi in this radially symmetrical charged distribution as
Phi=|vecE|cdot"Area of Gaussian Surface"=("Charge "Q_"eff")/epsilon_0

Inserting calculated values we get
|vecE|cdot4pir_1^2=(Q/(R^4)r_1^4)/epsilon_0
=>|vecE|=(Q/(R^4)r_1^4)1/(4pir_1^2)1/epsilon_0
=>|vec(E)| = (Qr_1^2)/(4piepsilon_0R^4)