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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.]

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alok kumar singh | Contributor-Level 10

3 months ago

1.30

Take a long thin wire XY of uniform linear charge density $λ$ . Consider a point A at a perpendicular distance l from the midpoint O of the wire. 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 = $λ d x$

Electric field due to the piece,

dE = = $\frac{1}{4 π ε_{0}} \times \frac{λ d x}{{(A Z)}^{2}}$ = $\frac{1}{4 π ε_{0}} \times \frac{λ d x}{(l^{2} + x^{2})}$ since AZ = $l^{2} + x^{2}$

The electric field is resolved into two rectangular components. dE $\cos ? θ$ is the perpendicular component and dE $\sin ? θ$ When the whole wire is considered, the component dE $\sin ? θ i s t h e p a r a l l e l c o m p o n e n t .$ is cancelled. Only the perpendicular component dE $\cos ? θ$ affects point A.

Hence

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1.30

Q = $λ d x$

Electric field due to the piece,

dE = = $\frac{1}{4 π ε_{0}} \times \frac{λ d x}{{(A Z)}^{2}}$ = $\frac{1}{4 π ε_{0}} \times \frac{λ d x}{(l^{2} + x^{2})}$ since AZ = $l^{2} + x^{2}$

Hence, effective electric field at point A due to element dx is $d E_{1}$

$d E_{1} = \frac{1}{4 π ε_{0}} \times \frac{λ d x \cos ? θ}{(l^{2} + x^{2})}$ ………………….(1)

In , $? A Z O$ , $\tan ? θ = \frac{x}{l}$ ,x = $l \tan ? θ$ ……….(2)

On differentiating equation (2), we get

$\frac{d x}{d θ} = l {s e c}^{2} θ$ , dx = …………….(3)

From equation (2), we have

$x^{2}$ + $l^{2}$ = $l^{2} {t a n}^{2} θ + l^{2}$ = $l^{2} {s e c}^{2} θ$ …….(4)

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

$d E_{1} = \frac{1}{4 π ε_{0}} \times \frac{λ l {s e c}^{2} θ d θ \cos ? θ}{l^{2} {s e c}^{2} θ}$ = $\frac{1}{4 π ε_{0}} \times \frac{λ \cos ? θ d θ}{l}$ …….(5)

The wire is so long that $θ$ tends from $- \frac{π}{2}$ to $\frac{π}{2}$

By integrating equation (5), we get

$\int_{- \frac{π}{2}}^{\frac{π}{2}} d E_{1=}$ $\int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{1}{4 π ε_{0}} \times \frac{λ \cos ? θ d θ}{l})$

$E_{1} = \frac{1}{4 π ε_{0}} \frac{λ}{l} {[\sin ? θ]}_{- \frac{π}{2}}^{\frac{π}{2}}$

$E_{1} = \frac{1}{4 π ε_{0}} \frac{λ}{l} \times 2 = \frac{λ}{2 π ε_{0} l}$

Therefore the electric field due to long wire is $\frac{λ}{2 π ε_{0} l}$

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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.]

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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.]

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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.]