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Obtain the Formula for the Electric Field Due to a Long Thin Wire of Uniform Linear Charge Density λ Without Using Gauss’S Law. - CBSE (Science) Class 12 - Physics

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Question

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.]

Solution

Take a long thin wire XY (as shown in the figure) of uniform linear charge density `lambda` .

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.

`therefore q=lambda dx`

Electric field due to the piece,

`dE=1/(4piin_0)(lambda dx)/((AZ)^2)`

However, `AZ=sqrt((l^2+x^2))`

`dE=(lambda dx)/(4piin_0(l^2+x^2))`

The electric field is resolved into two rectangular components. `dEcostheta` is the perpendicular component and  `dEsintheta` is the parallel component.

When the whole wire is considered, the component `dEsintheta` is cancelled.

Only the perpendicular component `dEcostheta` affects point A.

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

`therefore dE_1=(lambdadx costheta)/(4piin_0(x^2+l^2))`       ...(1)

In `triangleAZO,`

`tantheta=x/l`

`x=l tantheta`          ...(2)

On differentiating equation (2), we obtain `(dx)/(d theta)=lsec^2theta`  

`dx=lsec^2thata dtheta`

From equation (2),

`x^2+l^2=l^2+l^2 tan theta`

`therefore l^2(1+tan^2 theta)=l^2sec^2theta`

`thereforex^2+l^2=l^2sin^2theta`     ...(4)

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

`dE_1=(lambda l sec^2 d theta)/(4piin_0l^2sec^2 theta)xxcostheta`

`dE_1=(lambda cos theta d theta)/(4 pi in_0 l)`     .......(5)

The wire is so long that `theta` tends from `-pi/2`to `+pi/2`.

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

`int_(-pi/2)^(pi/2) dE_1=int_(-pi/2)^(pi/2)lambda/(4piin_0l) cos theta d theta`

`E_1=lambda/(4piin_0l)[sin theta]_(-pi/2)^(pi/2)`

`=lambda/(4 pi in_0l)xx2`

`E_1=lambda/(2piin_0l)`

Therefore, the electric field due to long wire is`lambda/(2piin_0l)`.

  Is there an error in this question or solution?

APPEARS IN

 NCERT Solution for Physics Textbook for Class 12 (2018 to Current)
Chapter 1: Electric Charge and Fields
Q: 30 | Page no. 49
Solution Obtain the Formula for the Electric Field Due to a Long Thin Wire of Uniform Linear Charge Density λ Without Using Gauss’S Law. Concept: Applications of Gauss’s Law.
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