Question

two point charges of 2 nano coulomb and 18 nc are located one meter apart in air. find the position along zhe line joining the two charges at which resultant electric intensity is zero​

Answer 1

Answer:

Given:

2 charges 2nC and 18 nC are located 1 metre apart in air along a straight line.

To find:

Position where the electric Field intensity is zero.

Concept :

Direction of electric Field intensity from a positive charge is away from the charge.

So when both the charges are positive , then there will be a point where the electric Field intensity from both the charges will cancel our each other.

That position is called the Nuetral Point.

Calculation :

Let that point be at a distance of x

from 2 nC charge.

So , distance of neutral point from 18nC will be (1 - x)

$\therefore \: E1 = E2$

$= > \dfrac{1}{4\pi \epsilon} \dfrac{2}{ {x}^{2} } = \dfrac{1}{4\pi \epsilon} \dfrac{18}{ {(1 - x)}^{2} }$

$= > \dfrac{1}{ {x}^{2} } = \dfrac{9}{ {(1 - x) }^{2} }$

Taking square root on both sides:

$= > \dfrac{1}{x} = \dfrac{3}{(1 - x)}$

$= > 1 - x = 3x$

$= > 4x = 1$

$= > x = \dfrac{1}{4} = 0.25 \: m$

So the neutral point is located at a distance of 0.25 m from 2nC and 0.75 m from 18nC.

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

For Point A

• Distance = x
• Charge = 2nC

For Point B

• Distance = x - 1
• Charge = 18 nC

As formula for Electric Intensity is :

$\huge {\boxed{\boxed{\rm{E \: = \: \dfrac{k Q}{d^2}}}}}$

Put E1 = E2

So,

$\implies {\sf{\dfrac{kQ_1}{x^2} \: = \: \dfrac{kQ_2}{(x \: - \: 1)^2}}} \\ \\ \implies {\sf{\dfrac{Q_1}{x^2} \: = \: \dfrac{Q_2}{(x \: - \: 1)^2}}} \\ \\ \implies{\sf{\dfrac{2}{x^2} \: = \: \dfrac{18}{(1 \: - \: x)^2}}} \\ \\ \implies {\sf{\dfrac{1}{x^2} \: = \: \dfrac{9}{((1 \: - \: x) ^2}}} \\ \\ \small {\underline{\pink{\sf{\: \: \: \: \: \: \: \dag \: Root \: Both \: Sides \: \: \: \: \: \: \: \:}}}} \\ \\ \implies {\sf{\dfrac{1}{x} \: = \: \dfrac{3}{x \: - \: 1}}} \\ \\ \implies {\sf{1 \: - \: x \: = \: 3x}} \\ \\ \implies {\sf{1 \: = \: 4x}} \\ \\ \implies {\sf{x \: = \: \dfrac{1}{4}}} \\ \\ {\underline{\sf{\therefore \: x \: is \: 0.24 \: m}}}$