Infinite charges of magnitude q each are lyin | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(a) Net field at origin $E = \frac{q}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{4^2}}} + ....\infty } \right]

Vedclass · Accepted Answer

$12 	imes 10^9 q N/C$

Vedclass · Answer

(a) Net field at origin $E = \frac{q}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{4^2}}} + ....\infty } ight]$
$ = \frac{q}{{4\pi {\varepsilon _0}}}\left[ {1 + \frac{1}{4} + \frac{1}{{16}} + .....\infty } ight]$
$ = \frac{q}{{4\pi {\varepsilon _0}}}\left[ {\frac{1}{{1 - \frac{1}{4}}}} ight] = 12 	imes {10^9}q\,N/C$

Infinite charges of magnitude $q$ each are lying at $x =1,\, 2,\, 4,\, 8...$ meter on $X$-axis. The value of intensity of electric field at point $x = 0$ due to these charges will be

Similar Questions

For a uniformly charged ring of radius $R$, the electric field on its axis has the largest magnitude at a distance $h$ from its centre. Then value of $h$ is

Equal charges $q$ are placed at the vertices $A$ and $B$ of an equilateral triangle $ABC$ of side $a$. The magnitude of electric field at the point $C$ is

The bob of a simple pendulum has mass $2\,g$ and a charge of $5.0\,\mu C$. It is at rest in a uniform horizontal electric field of intensity $2000\,\frac{V}{m}$. At equilibrium, the angle that the pendulum makes with the vertical is (take $g = 10\,\frac{m}{{{s^2}}}$)

What is the magnitude of a point charge which produces an electric field of $2\, N/coulomb$ at a distance of $60\, cm$ $(1/4\pi {\varepsilon _0} = 9 \times {10^9}\,N - {m^2}/{C^2})$

The charge per unit length of the four quadrant of the ring is $2\ \lambda , - 2\ \lambda , \lambda$ and $- \lambda$ respectively. The electric field at the centre is