Question

A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find the coordinates of the point where the net electric field strength due to these charges is zero.

Answer 1

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin

Second charge $q_{2}=6\times10^{-6}\ C$

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}$

$\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}$

$\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1$

$\dfrac{1}{d}=1.82$

$d=\dfrac{1}{1.82}$

$d=0.55\ m$

Hence, The coordinates of the point is (0,0.55).

Answer 2

It is a physical field occupied by a charged particle on another particle in its surrounding. The coordinates of the point where the net electric field strength due to these charges is zero will be (0,0.55).

What is an electric field?

It is a physical field occupied by a charged particle on another particle in its surrounding.

The following data are given as

q₁ is the first charge  at the origin

q₂ is the Second charge  

point of the Second charge is P = (0,1)

As we know the net electric field between the charges is zero at the midpoint.

The relation of the electric field with the distance between the charged particle is given by the formula

$\rm E=\frac{Kq}{d^2} \\\\0=\frac{K\times9\times10^{-6}}{d^2} +\frac{K\times6\times10^{-6}}{(1-d)^2} \\\\\frac{1-d}{d} =\sqrt{\frac{6}{9} } \\\\\frac{1}{d} =\sqrt{\frac{{6} }{9} } +1\\\\\rm\frac{1}{d} =1.82\\\\\rm d=0.55m$

Hence the coordinates of the point where the net electric field strength due to these charges is zero will be (0,0.55).

To learn more about the electric force refer to the link;

brainly.com/question/1076352