Two long, straight, parallel wires are 4 cm apart in air and carry uniform linear charge densities of \( +2\,\mu\mathrm{C\,m^{-1}} \) and \( -2\,\mu\mathrm{C\,m^{-1}} \). The magnitude of the electric field at the point midway between the wires is (Take \( \frac{1}{4\pi\varepsilon_0}=9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 3.6\times10^{6}\,\mathrm{N\,C^{-1}} \)
Each wire acts as an infinitely long line charge. Its field has the magnitude \( \lambda/2\pi\varepsilon_0 r \). The field points away from the wire for positive \( \lambda \) and towards it for negative \( \lambda \).
At the midpoint, the field of the positive wire points towards the negative wire.
The field of the negative wire also points towards that wire. So the two fields act in the same direction, and their magnitudes add.
The midpoint is at \( r=2\,\mathrm{cm}=0.02\,\mathrm{m} \) from each wire. Using \( \frac{1}{2\pi\varepsilon_0}=2\times\frac{1}{4\pi\varepsilon_0} \), the field of each wire there is:
\( \begin{aligned} E_1 &= \frac{2\times9\times10^{9}\times\lambda}{r} \\ &= \frac{2\times9\times10^{9}\times2\times10^{-6}}{0.02}\,\mathrm{N\,C^{-1}} \\ &= 1.8\times10^{6}\,\mathrm{N\,C^{-1}} \end{aligned} \)
The second wire has a charge density of the same magnitude, at the same distance. So \( E_2=1.8\times10^{6}\,\mathrm{N\,C^{-1}} \). The resultant field is:
\( \begin{aligned} E &= E_1+E_2 \\ &= 3.6\times10^{6}\,\mathrm{N\,C^{-1}} \end{aligned} \)
NCERT (2026-27):
Field due to an Infinitely Long Straight Uniformly Charged Wire – “Vectorially, E at any point is given by”
\( \displaystyle \vec{E}=\frac{\lambda}{2\pi\varepsilon_0 r}\,\hat{n} \)\( \qquad (1.32) \)
“where \( \hat{n} \) is the radial unit vector in the plane normal to the wire passing through the point. E is directed outward if λ is positive and inward if λ is negative.”