Derive the expression of ELECTROSTATIC FIELD INTENSITY for an infinitely long wire containing charge "q".
Show the solution without using Gauss's Law.
#Revision Q12
Answers
- Derive the expression of Electrostatic Field Intensity for an infinitely long wire containing charge
- Expression of Electrostatic Field Intensity for an infinitely long wire containing charge =
Consider the figure below, (Refer to Attachment)
- Consider a small element on the wire of length . Charge on the element is,
Electric field due to the small element is,
Total electric field at the given point is only due to perpendicular components since parallel components get cancelled. Therefore,
First Let us Simplify
- Compute the Indefinite Integral
Apply Trigonometric Substitution :
So dx =
Take the Constant Out :
Use the following identity :
Use the common integral :
Substitute back
Use the following identity :
Add a Constant to the Solution
- Compute the Boundaries
EXPLANATION.
Electric field intensity for an infinitely long wire containing charge ''q''.
As we know that, the formula.
⇒ Eₓ = kλ/d[Sin∅₁ + Sin∅₂].
⇒ Ey = kλ/d[Cos∅₁ - Cos∅₂].
As we know that,
In infinitely long wire ∅ varies from ∞ to ∞.
⇒ ∅₁ = 90°
⇒ ∅₂ = 90°.
E(net) = [E₁ + E₂].
⇒ Eₓ = kλ/d [ Sin90° + Sin90°].
⇒ Eₓ = kλ/d [1 + 1].
⇒ Eₓ = 2kλ/d.
⇒ Ey = kλ/d [ Cos90° - Cos90°].
⇒ Ey = kλ/d [ 0 - 0 ].
⇒ Ey = 0.
⇒ E(net) = E₁ + E₂.
⇒ E(net) = 2kλ/d + 0.
⇒ E(net) = 2kλ/d.
As we know that,
⇒ k = 1/4πε°.
⇒ E(net) = 2λ/4πε°d.
⇒ E(net) = λ/2πε°d.