An infinite number of electric charges, each | vedclass

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

Question

(C) The electric field intensity due to a point charge is given by $E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}$.Since the consecutive charg

Vedclass · Accepted Answer

$36 	imes 10^4$

Vedclass · Answer

(C) The electric field intensity due to a point charge is given by $E = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}$.Since the consecutive charges are of opposite signs, the net electric field at $x = 0$ is the sum of fields due to individual charges:$E = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{Q}{r_1^2} - \frac{Q}{r_2^2} + \frac{Q}{r_3^2} - \frac{Q}{r_4^2} + \dots \infty ight]$$E = \frac{Q}{4 \pi \varepsilon_0} \left[ \frac{1}{(1 	imes 10^{-2})^2} - \frac{1}{(2 	imes 10^{-2})^2} + \frac{1}{(4 	imes 10^{-2})^2} - \frac{1}{(8 	imes 10^{-2})^2} + \dots \infty ight]$$E = (9 	imes 10^9) 	imes (5 	imes 10^{-9}) 	imes 10^4 \left[ \frac{1}{1^2} - \frac{1}{2^2} + \frac{1}{4^2} - \frac{1}{8^2} + \dots \infty ight]$$E = 45 	imes 10^4 \left[ 1 - \frac{1}{4} + \frac{1}{16} - \frac{1}{64} + \dots \infty ight]$This is an infinite geometric progression with first term $a = 1$ and common ratio $r = -\frac{1}{4}$.The sum $S_{\infty} = \frac{a}{1 - r} = \frac{1}{1 - (-1/4)} = \frac{1}{5/4} = \frac{4}{5}$.Therefore, $E = 45 	imes 10^4 	imes \frac{4}{5} = 36 	imes 10^4 	ext{ N/C}$.

Similar Questions

$A$ charged water drop of radius $0.1 \, \mu m$ is in equilibrium under an electric field. The charge on the drop is equivalent to the electronic charge. The intensity of the electric field is ........ $N/C$.

The electric field at the centre $O$ of a semicircle of radius $a$ having a linear charge density $\lambda$ is given by:

Two charges $+q$ and $-q$ are situated at a certain distance. At the point exactly midway between them:

Two charges $Q$ and $-3Q$ are placed at a certain distance $x$. If the electric field at the position of charge $Q$ due to charge $-3Q$ is $E$,then what is the electric field at the position of charge $-3Q$ due to charge $Q$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module