Question

If the $\lambda =1\mu C/m,$ then electric field intensity at O is:

$\begin{array}{rl}& A.9N/C\\ & B.900N/C\\ & C.9000N/C\\ & D.9\times {10}^{9}N/C\end{array}$

Answer 1

Hint: In order to find the solution of this equation we will use the formula related linear charge density ($\lambda$) and the electric field density and that is $E={\displaystyle \frac{2k\lambda }{r}}$ by using this equation we will get electric field intensity at point o.

Formula used:
$E={\displaystyle \frac{2k\lambda }{r}}$
E = electric field intensity
$\lambda$= linear charge density
r = distance r from the line

Complete step-by-step answer:
Now it is given that in the question that the value of the linear charge density is
$\lambda =1\mu C/m$
$\to$ So if we convert from $\mu C/m$ to C/m we have to multiply by ${10}^{-6}$ now the value of the linear charge density is
$\lambda =1\times {10}^{-6}C/m$

$\to$Now the electric field intensity at the distance from 1m is given by
$E_1={\displaystyle \frac{2k\lambda }{r}}$
Here the value of k is given by $9\times {10}^9$
$\begin{array}{rl}& E_1=\underset{1}{2\times 9\times {10}^9\times 1\times {10}^{-6}}\\ & E_1=18\times {10}^{3}N/C......\left(1\right)\end{array}$
$\to$Now the electric field intensity at distance from 2m is given by
$\begin{array}{rl}& E_1=\underset{2}{2\times 9\times {10}^9\times 1\times {10}^{-6}}\\ & E_1=9\times {10}^{3}N/C......\left(2\right)\end{array}$
$\to$Now the total electric field intensity at point o is given by
$E=E_1-E_2......\left(3\right)$
$\to$Now substitute the value of the equation (1) and (2) in equation (3) to get electric field intensity at point o.
$\begin{array}{rl}& E=18\times {10}^3-9\times {10}^3\\ & =(18-9)\times {10}^3\\ & =9\times {10}^{3}N/C\\ & E=9000N/C\end{array}$
Hence the correct option is (C) 9000N/C

So, the correct answer is “Option C”.

Additional Information: In this question value of the k is given by ${\displaystyle \frac{1}{4\pi {\epsilon }_0}}$ because the initial equation of the electric field intensity is
$E={\displaystyle \frac{\lambda }{2\pi {\epsilon }_{0}r}}$
Then we can substitute${\displaystyle \frac{1}{2\pi {\epsilon }_0}}$ by
$\begin{array}{rl}& 2k=2\left({\displaystyle \frac{1}{4\pi {\epsilon }_0}}\right)\\ & 2k={\displaystyle \frac{1}{2\pi {\epsilon }_0}}\end{array}$
Hence our equation will become
$E={\displaystyle \frac{2k\lambda }{r}}$
Hence the correct option is (C).

Note: As shown in the figure that when the positive (+ve) charge is considered at a point o then from the both the sides the linear charge density will be opposite from the both the sides at the point o.