Question #135892
State the laws of electrostatics in integral form.
(b) A uniformly charged long cylinder of radius a and length L has
total charge q inside its volume.
What is the direction of the electric field at points outside the cylinder?
Find the electric field inside and outside the cylinder.
1
Expert's answer
2020-10-01T10:11:05-0400

(a) SEˉdSˉ=qϵ0\displaystyle \oint_S \bar{E}\cdot \bar{dS}=\frac{q}{\epsilon_0} (Gauss law in SI)

CEˉdlˉ=0\displaystyle \oint_C \bar{E}\cdot \bar{dl}=0 (conservative filed law)

(b) Due to cylindrical symmetry, the electric field will be pointed from the axis of symmetry to outside. Let's use Gauss law.

(i) inside the cylinder (r<a)

0aρ(r)dV=q\displaystyle \int_0^a \rho(r) dV =q (in volume πa2L\pi a^2L )

E2πrL=1ϵ00rρ(r)dV=1ϵ0qπr2Lπa2L=1ϵ0qr2a2\displaystyle E \cdot 2\pi r L = \frac{1}{\epsilon_0} \int_0^r \rho(r) dV = \frac{1}{\epsilon_0} q \frac{\pi r^2 L}{\pi a^2 L}= \frac{1}{\epsilon_0} q \frac{r^2}{a^2}

E=qrϵ02πLa2\displaystyle E = \frac{qr}{\epsilon_0 \, 2 \pi L a^2}

(ii) outside of the cylinder (r>a)

E2πrL=1ϵ00aρ(r)dV=1ϵ0q\displaystyle E \cdot 2\pi r L = \frac{1}{\epsilon_0} \int_0^a \rho(r) dV = \frac{1}{\epsilon_0} q

E=qϵ02πLr\displaystyle E = \frac{q}{\epsilon_0 \,2 \pi L r}

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