Class-12th Physics, CBSE and UP Board

Electric Charges and Fields,

Chapter-1, Exercise -1, Additional Exercise, Q-1.30

NCERT Solutions for Class 12th Physics

# 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.]

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Take a long thin wire XY of uniform linear charge density λ.

Consider a point A at a perpendicular distance

lfrom the mid-point 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.

Therefore q =λdx

Electric field due to the piece,

dE = 1 /4πε_{0}. λdx/(AZ)²However , AZ = √

(l²+ x²)Thereforeis the parallel component. When the whole wire is considered, the component

dE = 1 /4πε_{0}. λdx/(l²+ x ²)The electric field is resolved into two rectangular components.

dEcosθis the perpendicular component anddEsinθdEsinθis cancelled. Only the perpendicular componentdEcosθaffects point A. Hence, effective electric field at point A due to the elementdxisdE₁.Therefore

dE₁ =1 /4πε_{0}. λdx cosθ/(l²+ x ²)—–Eq-1In ΔAZO,

tanθ=x/l ⇒ x = l.tan θ —————————–Eq-2From Equation-2 we obtain

dx/dθ = l sec²θ ⇒ dx= l sec²θ dθ ——————————Eq-3From Equation-2 we have

x² + l² = l² tan² θ +l²= l² ( tan² θ +1)= l² sec²θ———-Eq-4Putting equations-3 & 4 in Equation-1 ,we obtain

dE₁ =1 /4πε_{0}. λ (l sec²θ dθ ) cosθ/(l² sec²θ)=

1 /4πε_{0}.λcosθ dθ /l—————————-Eq-5The wire is so long that

θ tends from -π/2 to π/2By integrating Eq-5 ,we obtain the value of field

E₁ as,⌠

^{π/2 }dE ₁ = ⌠^{π/2 }1 /4πε_{0 . }λcosθ dθ /l_{⁻ π/2}⌡_{⁻ π/2}⌡⇒ E₁ =1 /4πε_{0 . }λ/l [sinθ_{⁻ π/2 }]^{π/2}⇒ E₁ =1 /4πε_{0 . }λ/l [sinθ_{⁻ π/2 }]^{π/2}⇒ E₁ =1 /4πε_{0 . }λ/lx 2⇒ E₁ = λ/2πε_{0 }lTherefore .the electric field due to long wire is

λ/2πε_{0 }l