Question

derive an expression for a wire from infinity have line charge density Lemda without using gauss law...​

Answer 1

Let a infinite long thin wire of surface charge density \lambdaλ  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}dEp=4πϵ0r2Q  

we know, charge = lines charge density × length 

So, Q=\lambda dyQ=λdy 

r² = y² + a² [ from Pythagoras theorem, ]

\implies dE_P=\frac{\lambda dy}{4\pi\epsilon_0(y^2 + a^2)}⟹dEP=4πϵ0(y2+a2)λdy 

E_p=\int\limits^{\infty}_{-\infty}{dE_p}Ep=−∞∫∞dEp 

here you can see that vertical components cancel out and horizontal component are added due to symmetry. 

\implies E_P=\int\limits^{\infty}_{0}{2dE_pcos\theta}⟹EP=0∫∞2dEpcosθ 

\implies E_p=\int\limits^{\infty}_{0}{\frac{\lambda dy}{4\pi\epsilon_0(y^2+a^2)}cos\theta}⟹Ep=0∫∞4πϵ0(y2+a2)λdycosθ 

=\frac{2}{4\pi\epsilon_0}\int\limits^{\infty}_0{\frac{\lambda dy}{y^2+a^2}}\times\frac{y}{\sqrt{y^2+a^2}}=4πϵ020∫∞y2+a2λdy×y2+a2y  

=\frac{1}{2\pi\epsilon_0}\int\limits^{\infty}_0{\frac{\lambda ydy}{(y^2+a^2)^{3/2}}}=2πϵ010∫∞(y2+a2)3/2λydy 

take y² + a² = x 

differentiate with respect to x

2ydy = dx 

and taking proper limits

I mean, take upper limit y = ∞² + a² = ∞

and lower limit , y = 0² + a² = a²

E_p=\frac{\lambda}{4\pi\epsilon_0}\int\limits^{\infty}_{a^2}{\frac{dx}{x^{3/2}}}Ep=4πϵ0λa2∫∞x3/2dx 

\implies\left[\begin{array}{c}\frac{-2\lambda}{4\pi\epsilon_0}\frac{1}{\sqrt{x}}\end{array}\right]^{\infty}_{a^2}= \frac{\lambda}{2\pi\epsilon_0 a}⟹[4πϵ0−2λx1]a2∞=2πϵ0aλ 

hence, E=\frac{\lambda}{2\pi\epsilon_0 a}E=2πϵ0a

Answer 2

Answer:

Take a long thin wire of uniform linear charge density λ.

Consider a point A at a perpendicular distance l from the mid-point O of the wire, as shown in the following figure.

Let E be the electric field at point A due to the wire, XY.

Consider a small length element dx on the wire section with OZ=x

Let q be the charge on this piece.

∴q=λdx

Electric field due to the piece,

dE=4π∈01(AZ)2λdx

However, AZ=(l2+x2)

∴dE=4π∈0(l2+x2)λdx

The electric field is resolved into two rectangular components. dEcosθ is the perpendicular component and dEsinθ is the parallel component.

When the whole wire is considered, the component dEsinθ is cancelled.

Only the perpendicular component dEcosθ affects point A.

Hence, effective electric field at point A due to the element dx is dE1.

∴dE1=4π∈0(x2+l2)λdxcosθ  ... (1)

In △AZO,

tanθ=lx

x=ltanθ  ... (2)

On differentiating equation (2), we obtain

dx=lsec2(θ)dθ

dE1