Question

a ring of radius R has uniform linear charge density Lambda the electric field at the center of the radius is. a)0. b)1/2πe0 c)1/2πe0 lambda /r^2​

Answer 1

Answer:

2kλ/R

Explanation

use the formula of length of the arc

that is = πr/2

charge on left of =λπr/2=Q

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Answer 2

The length of arc with a positive charge$=\frac{2\pi R}{4} =\frac{\pi }{2}$

$\therefore$ Charge on left half $=\frac{\lambda \pi R}{2}=Q$

The electric field due to the circular arc is given by

$E=\frac{2kQ sin(\frac{\phi}{2})}{\phi R ^2}$ (here $\phi =\frac{\pi }{2}$ )

putting values we get

$E_1=\frac{2k\lambda \pi R \times 2}{2\sqrt{2}\pi R^2 }$

$E_1=\frac{\sqrt{2} K\lambda}{R}$

Similar will be electric field due to right half but in direction perpendicular, as shown in figure

$\therefore E_{net}=\sqrt{(\frac{\sqrt{2}K\lambda }{R} )^2+(\frac{\sqrt{2}K\lambda }{R} )^2}$

$E_{net}=\frac{2k\lambda}{R}$

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