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Question

(B) The electric field $E$ due to an infinitely long straight wire with linear charge density $\lambda$ at a distance $r$ is given by $E = \frac{\lamb

Vedclass · Accepted Answer

$\frac{\lambda}{\pi \varepsilon_0 R}$

Vedclass · Answer

(B) The electric field $E$ due to an infinitely long straight wire with linear charge density $\lambda$ at a distance $r$ is given by $E = \frac{\lambda}{2 \pi \varepsilon_0 r}$.For the first wire with charge density $\lambda_1 = 2 \lambda$,the electric field at the midpoint (distance $r = R/2$) is $E_1 = \frac{2 \lambda}{2 \pi \varepsilon_0 (R/2)} = \frac{2 \lambda}{\pi \varepsilon_0 R}$.For the second wire with charge density $\lambda_2 = 3 \lambda$,the electric field at the midpoint (distance $r = R/2$) is $E_2 = \frac{3 \lambda}{2 \pi \varepsilon_0 (R/2)} = \frac{3 \lambda}{\pi \varepsilon_0 R}$.Since the wires are parallel and positively charged,the fields at the midpoint point in opposite directions.The net electric field intensity is $E_{net} = |E_2 - E_1| = |\frac{3 \lambda}{\pi \varepsilon_0 R} - \frac{2 \lambda}{\pi \varepsilon_0 R}| = \frac{\lambda}{\pi \varepsilon_0 R}$.

Two infinitely long thin straight wires having uniform linear charge densities $2 \lambda$ and $3 \lambda$ are arranged parallel to each other at a distance $R$ apart. The intensity at a point midway between them is

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