Question

Two point charges of 20 micro coulomb and 80 micro coulomb are placed 18 cm apart. Find the position where the electric field is zero?

Answer 1

Given:

Point charges 20 micro-Coulomb and 80 micro-Coulomb are placed 18 cm apart.

To find:

Position at which the net Electrostatic Field Intensity is zero.

Concept:

The required position , where the Electrostatic Field Intensity is zero is known as the Neutral Point.

If any charge is placed at that position , it won't experience any force.

Calculation:

Let the position of zero Electrostatic Field be located x cm distance from 20 $\mu C$.

So we can say that the position of that neutral point will be (18-x) cm from the other charge .

At the neutral point , the Electrostatic Field intensity of one charge will be nullified by the electrostatic field intensity of the other charge.

This is because the field intensities are equal and opposite ;

$\therefore E_{1} = E_{2}$

$= > \dfrac{1}{4\pi\epsilon_{0}} \dfrac{20 \times {10}^{ - 6} }{ {x}^{2} } = \dfrac{1}{4\pi\epsilon_{0}} \dfrac{80 \times {10}^{ - 6} }{ {(18 - x)}^{2} }$

Cancelling the common terms :

$= > \cancel{ \dfrac{1}{4\pi\epsilon_{0}}} \: \dfrac{20 \times \cancel{{10}^{ - 6}} }{ {x}^{2} } = \cancel{\dfrac{1}{4\pi\epsilon_{0}}} \: \dfrac{80 \times \cancel{{10}^{ - 6}} }{ {(18 - x)}^{2} }$

$= > \dfrac{20}{ {x}^{2} } = \dfrac{80}{ {(18 - x)}^{2} }$

$= > \dfrac{ {x}^{2} }{ {(18 - x)}^{2} } = \dfrac{20}{80}$

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

Taking Square root on both sides :

$= > \dfrac{x}{18 - x} = \dfrac{1}{2}$

$= > 2x = 18 - x$

$= > 3x = 18$

$= > x = 6 \: cm$

So the neutral point is located 6 cm from 20 micro-Coulomb charge and 12 cm from 80 micro-Coulomb charge.