Question 1.30) Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E without using Gausss law

Consider a long thin wire of uniform linear charge density,

λ

.

Formula for electric field due to this wire at any point P at a perpendicular distance PC = r from the wire.

Consider a small element of length dx of the wire with centre O, such that OC = x.

Charge on the element, q =

λ . dx

Electric intensity at P due to the element is given by,

dE = \frac{1}{4 {πε}_{o}} \frac{λ . dx}{{OP}^{2}} = \frac{λ . dx}{4 {πε}_{o} (r^{2} + x^{2})}

Now,

\overset{⇀}{dE}

can be resolved into two rectangular components, that is

\overset{⇀}{dE} \cos θ

in a perpendicular direction and

\overset{⇀}{dE} \sin θ

in a parallel direction.

The parallel component will be cancelled by the parallel component of the field due to charge on a similar element dx of wire on the other half.

Effective component of electric intensity due to the charge element,

dE' = \overset{⇀}{dE} \cos θ

dE' = \frac{λ . dx \cos θ}{4 {πε}_{o} (r^{2} + x^{2})}

...(i)

From

∆ OCP, x = r \tan θ

∴ dx = r \sec^{2} θ dθ

Now, r^{2} + x^{2} = r^{2} + r^{2} \tan^{2} θ = r^{2} (1 + \tan^{2} θ) = r^{2} \sec^{2} θ

From equation (i), we have

dE' = \frac{λ r \sec^{2} θ dθ}{4 {πε}_{o} r^{2} \sec^{2} θ} \cos θ

dE' = \frac{λ}{4 {πε}_{o} r} \cos θ dθ

Since the wire has infinite length, it’s ends A and B are infinite distances apart.

θ

varies from

- \frac{π}{2} to + \frac{π}{2}

Electric Intensity at P due to the whole wire is given by,

E' = \int_{- \frac{π}{2}}^{+ \frac{π}{2}} \frac{λ}{4 {πε}_{o} r} \cos θ dθ

E' = \frac{λ}{4 {πε}_{o} r} {[\sin θ]}_{- \frac{π}{2}}^{+ \frac{π}{2}}

= \frac{λ}{2 {πε}_{o} r}

, is the required electric field intensity.

Question 1.30) Obtain the formula for the electric field due to a long thin wire of uniform linear charge density E without using Gausss law - Hosting Precise

Posted on 9th Oct 2021 09:26:27 in

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

Ans:-

Consider a long thin wire of uniform linear charge density, $λ$ .

Formula for electric field due to this wire at any point P at a perpendicular distance PC = r from the wire.

Consider a small element of length dx of the wire with centre O, such that OC = x.

Charge on the element, q = $λ . dx$

Electric intensity at P due to the element is given by,

$dE = \frac{1}{4 {πε}_{o}} \frac{λ . dx}{{OP}^{2}} = \frac{λ . dx}{4 {πε}_{o} (r^{2} + x^{2})}$

Now, $\overset{⇀}{dE}$ can be resolved into two rectangular components, that is $\overset{⇀}{dE} \cos θ$ in a perpendicular direction and $\overset{⇀}{dE} \sin θ$ in a parallel direction.

The parallel component will be cancelled by the parallel component of the field due to charge on a similar element dx of wire on the other half.

Effective component of electric intensity due to the charge element, $dE' = \overset{⇀}{dE} \cos θ$

$dE' = \frac{λ . dx \cos θ}{4 {πε}_{o} (r^{2} + x^{2})}$ ...(i)

From $∆ OCP, x = r \tan θ$

$∴ dx = r \sec^{2} θ dθ$

$Now, r^{2} + x^{2} = r^{2} + r^{2} \tan^{2} θ = r^{2} (1 + \tan^{2} θ) = r^{2} \sec^{2} θ$

From equation (i), we have

$dE' = \frac{λ r \sec^{2} θ dθ}{4 {πε}_{o} r^{2} \sec^{2} θ} \cos θ$

$dE' = \frac{λ}{4 {πε}_{o} r} \cos θ dθ$

Since the wire has infinite length, it’s ends A and B are infinite distances apart.

$θ$ varies from $- \frac{π}{2} to + \frac{π}{2}$

Electric Intensity at P due to the whole wire is given by,

$E' = \int_{- \frac{π}{2}}^{+ \frac{π}{2}} \frac{λ}{4 {πε}_{o} r} \cos θ dθ$

$E' = \frac{λ}{4 {πε}_{o} r} {[\sin θ]}_{- \frac{π}{2}}^{+ \frac{π}{2}}$

$= \frac{λ}{2 {πε}_{o} r}$ , is the required electric field intensity.

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