An infinite number of electric charges,each w | vedclass

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

Question

(C) The electric field $E$ at the origin due to a point charge $q$ at distance $r$ is given by $E = k \frac{q}{r^2}$,where $k = 9 	imes 10^9 \, N \cd

Vedclass · Accepted Answer

$36 	imes 10^4$

Vedclass · Answer

(C) The electric field $E$ at the origin due to a point charge $q$ at distance $r$ is given by $E = k \frac{q}{r^2}$,where $k = 9 	imes 10^9 \, N \cdot m^2/C^2$.Since the charges alternate in sign,the field at $x = 0$ is:$E = k \cdot q \left[ \frac{1}{(1 	imes 10^{-2})^2} - \frac{1}{(2 	imes 10^{-2})^2} + \frac{1}{(4 	imes 10^{-2})^2} - \frac{1}{(8 	imes 10^{-2})^2} + \dots ight]$$E = (9 	imes 10^9) 	imes (5 	imes 10^{-9}) 	imes 10^4 \left[ 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \dots ight]$$E = 45 	imes 10^4 \left[ 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \dots ight]$This is a geometric series with first term $a = 1$ and common ratio $r = -1/4$.The sum $S = \frac{a}{1 - r} = \frac{1}{1 - (-1/4)} = \frac{1}{5/4} = \frac{4}{5}$.$E = 45 	imes 10^4 	imes \frac{4}{5} = 9 	imes 10^4 	imes 4 = 36 	imes 10^4 \, N/C$.

Similar Questions

In the Millikan's experiment,the distance between two horizontal plates is $2.5 \, cm$ and the potential difference applied is $250 \, V$. The electric field between the plates will be ....... $V/m$.

Find out the electric field intensity at point $A(1, 0, 2)$ due to a point charge $-20\,\mu C$ situated at point $B(0, 2, 1)$.

What is the magnitude of a point charge which produces an electric field of $2 \, N/C$ at a distance of $60 \, cm$? (Given: $1/(4\pi \varepsilon_0) = 9 \times 10^9 \, N \cdot m^2/C^2$)

Two charges $Q$ and $4Q$ are separated by a distance of $6 \text{ cm}$. The distance of the point from $4Q$ at which the net electric field is zero is: (in $\text{ cm}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module