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NCERT Class XII Chapter
Electric Charges and Fields
Questions With Solutions

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Question : 30
Total: 34
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss’s law.
Solution:  
Consider a point P, a unit away from the long charged wire. Electric field due to element dy,

dEp =
Q
4πε0r2
 or Q = λdy
r2 = y2+a2
⇒ dEp =
λdy
4πε0(y2+a2)
 ⇒ Ep =
∞
∫
−∞
 dEp
Vertical components cancel out and horizontal components are added due to symmetry
⇒ Ep =
∞
∫
0
2λdy
4πε0(y2+a2)
 × (cosθ)1
=
2
4πε0
‌
∞
∫
0
λdy
y2+a2
 ×
y
√y2+a2
 =
2
4πε0
‌
∞
∫
0
λydy
(y2+a2)3∕2
Take y2+a2 = x ⇒ 2ydy = dx
Taking proper limits
Ep =
λ
4πε0
‌
∞
∫
a2
dx
x3∕2
⇒ [
−2λ
4πε0
1
√x
]a∞2 =
λ
2πε0a
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