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Question

(B) The electric field $E$ due to an infinitely long wire with linear charge density $\lambda$ at a distance $r$ is given by $E = \frac{\lambda}{2 \pi

Vedclass · Accepted Answer

$36 	imes 10^{11} \ NC^{-1}$

Vedclass · Answer

(B) The electric field $E$ due to an infinitely long wire with linear charge density $\lambda$ at a distance $r$ is given by $E = \frac{\lambda}{2 \pi \epsilon_0 r} = \frac{2k\lambda}{r}$,where $k = 9 	imes 10^9 \ Nm^2C^{-2}$.Given $\lambda_1 = 4 \ Cm^{-1}$,$\lambda_2 = 8 \ Cm^{-1}$,and separation $d = 4 \ cm = 0.04 \ m$.The midpoint is at distance $r = d/2 = 0.02 \ m$ from both wires.The electric field due to wire $1$ is $E_1 = \frac{2 	imes 9 	imes 10^9 	imes 4}{0.02} = 36 	imes 10^{11} \ NC^{-1}$ (directed away from wire $1$).The electric field due to wire $2$ is $E_2 = \frac{2 	imes 9 	imes 10^9 	imes 8}{0.02} = 72 	imes 10^{11} \ NC^{-1}$ (directed away from wire $2$).Since both wires have positive charge densities,the fields at the midpoint are in opposite directions.The net electric field is $E_{net} = |E_2 - E_1| = |72 	imes 10^{11} - 36 	imes 10^{11}| = 36 	imes 10^{11} \ NC^{-1}$.

The linear charge densities of two infinitely long thin and parallel wires are $4 \ Cm^{-1}$ and $8 \ Cm^{-1}$,and the separation between them is $4 \ cm$. The electric field intensity at the midpoint on the line joining them is:

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