Question

Derive electric field due to long thin wire witout gauss law

Answer 1

The ends of the cylinder do not contribute as the electric field cancels. Therefore the area A = 2*pi*r*L, where

r = radius in m from wire, either 0.005m or 0.015m for this problem

L = length of cylinder

Since you have charge/length this works nicely, as you can assume 1m length.

The only remaining consideration is the enclosed charge. When r=0.005m, you are inside the pipe and therefore the enclosed charge is 6.5nC/m. When r=0.015m, you are outside the pipe and the enclosed charge is equal to 6.5-3.2 = 3.3nC/m.

So for 0.005m,

E = 6.5e-9C/(2*pi*0.005m)

And for 0.015m,

E = 3.3e-9C/(2*pi*0.015m)

Answer 2

Explanation:

Brainly.in

What is your question?

R0HlT

Obtain the formula for electric fie

Let a infinite long thin wire of surface charge density \lambda is placed vertically as shown in figure. consider a point P, a unit away from the long charged wire. electric field due to element dy ,

dE_p=\frac{Q}{4\pi\epsilon_0r^2}

we know, charge = lines charge density × length

So, Q=\lambda dy

r² = y² + a² [ from Pythagoras theorem, ]

\implies dE_P=\frac{\lambda dy}{4\pi\epsilon_0(y^2 + a^2)}

E_p=\int\limits^{\infty}_{-\infty}{dE_p}

here you can see that vertical components cancel out and horizontal component