Let there be spherically symmetric charge distribution with charge density P(x)=Pₒ[(5/4)-(r/R)] up-to r=R, P(r)=0 for r≻R. Where 'r'is the distance from origin. What is the electric field at a distance 'r' (r≻R) form the origin?

2 Answers
Jun 10, 2018

=rho_o/(4epsilon_o)(5/3-r/R)

Explanation:

volume charge density, rho(r)=rho_o(5/4-r/R), upto r=R
=>(dq)/(dV)=rho_o(5/4-r/R)
=>dq=rho_o(5/4-r/R)dV...(i)

consider a sphere of radius R
PCPC
let a thin strip of radius dr
the surface area of the strip is 4pir^2 and thickness is dr
so, the small volume of the strip is dV=4pir^2dr
now
substitute for dV in (i)
dq=rho_o(5/4-r/R)4pir^2dr

now integrate to find out total charge enclosed till any point 'r'
=>q=intrho_o(5/4-r/R)4pir^2dr
=>q=rho_opir^3(5/3+r/R)
now electric field at ny point at a distance r is given by
E=kq/r^2=(krho_opir^3(5/3-r/R))/r^2
=rho_o/(4epsilon_o)(5/3-r/R)

Jun 10, 2018

E(r) = ( rho_o R^3)/(6 \ varepsilon_o ) * 1/r^2

Explanation:

  • rho(r) = {( rhoₒ(5/4-r/R), 0 le r le R),(0, r gt R):}

Gauss' Law states that:

  • The total of the electric flux Phi out of a closed surface is equal to the total charge enclosed divided by the permittivity:

  • Phi = Q_("enc")/varepsilon_o

If we draw a Gaussian sphere enclosing, and with the same origin as, this spherically symmetric charge distribution, then:

  • The total of the electric flux out of the Gaussian sphere is equal to E-field at the surface times the surface's area:

  • Phi = E(r) * 4 pi r^2.

IOW, for a Gaussian sphere , concentric with the charge distribution, and of radius r gt R:

  • E(r) = Q_("enc")/(4 pi varepsilon_o r^2) (Coulomb's Law with the distribution treated as a point charge)

(1) Find Q_("enc")

Q_("enc") = int_V rho(r) \ dV

For a sphere:

  • V = 4/3 pi r^3, qquad dV = 4 pi r^2 dr

Q_("enc") = 4 pi rho_o int_0^R (5/4-r/R) r^2 \ dr

= 2/3 pi rho_o R^3

(2) Find E(r), qquad r gt R

E(r) = Q_("enc")/(4 pi varepsilon_o r^2)

= ( rho_o R^3)/(6 \ varepsilon_o ) * 1/r^2