Question

Question 1.30: Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law. [Hint: Use Coulomb’s law directly and evaluate the necessary integral.]

Class 12 - Physics - Electric Charges And Fields Electric Charges And Fields Page-49

Answer 1

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Take a long thin wire XY (as shown in the following figure) 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,

However, 

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 .

∴ d E1 = 1/4πε0 x λdx. cos θ/(l2 + x2)                  ...(1)

In ∆AZO, tanθ = x/l ⇒ x = l.tanθ                          ...(2)

On differentiating equation (2), we obtain

dx/dθ = l x sec 2 θ ⇒ dx = l x sec 2 θdθ              ...(3)

From equation (2), we have

x2 + l2 = l2tan2θ + l2 = l2 (tan2 θ + 1) = l2 sec2θ          ...(4)

Putting equations (3) and (4) in equation (1), we obtain



The wire is so long that θ tends from − π/2 to π/2.

By integrating equation (5), we obtain the value of field E1 as,


 

Therefore, the electric field due to long wire is λ/2πε0 l .


IN HINDI


एक्स 2) ... (1)

ΔAZO में, tanθ = x / l ⇒ x = l.tanθ ... (2)

समीकरण (2) को अलग करने पर, हम प्राप्त करते हैं

डीएक्स / डीआरटी = एल एक्स सेकंड 2 θ ⇒ डीएक्स = एल एक्स सेकंड 2 θdθ ... (3)

समीकरण (2) से, हमारे पास है

x2 + l2 = l2tan2θ + l2 = l2 (tan2 θ + 1) = l2 sec2 θ ... (4)

समीकरण (3) और (4) समीकरण (1) में डालकर, हम प्राप्त करते हैं

तार इतना लंबा है कि θ - π / 2 से π / 2 तक रहता है।

समीकरण (5) को एकीकृत करके, हम फ़ील्ड ई 1 के मान को प्राप्त करते हैं,

 

इसलिए, लंबे तार के कारण बिजली क्षेत्र λ / 2πε0 एल है।

Answer 2

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 are added due to symmetry.
$\implies E_P=\int\limits^{\infty}_{0}{2dE_pcos\theta}$

$\implies E_p=\int\limits^{\infty}_{0}{\frac{\lambda dy}{4\pi\epsilon_0(y^2+a^2)}cos\theta}$

$=\frac{2}{4\pi\epsilon_0}\int\limits^{\infty}_0{\frac{\lambda dy}{y^2+a^2}}\times\frac{y}{\sqrt{y^2+a^2}}$

$=\frac{1}{2\pi\epsilon_0}\int\limits^{\infty}_0{\frac{\lambda ydy}{(y^2+a^2)^{3/2}}}$

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

$E_p=\frac{\lambda}{4\pi\epsilon_0}\int\limits^{\infty}_{a^2}{\frac{dx}{x^{3/2}}}$

=> $\left[\begin{array}{c}\frac{-2\lambda}{4\pi\epsilon_0}\frac{1}{\sqrt{x}}\end{array}\right]^{\infty}_{a^2}= \frac{\lambda}{2\pi\epsilon_0 a}$

hence,E= $\frac{\lambda}{2\pi\epsilon_0a}$