A point charge +6q is located at the origin, and a point charge -4q is located on the x-axis at D = 0.530 m. At what location on the x-axis will a third charge, q0, experience no net force from the other two charges?

Explanation:

A point charge (+6q) at the origin I.e at (0,0)

Another point(-4q) charge is located at (0.53m, 0)

Both charges are on the x axis

Where will a third charge be place and it will experience no net force?

Fnet=0 due to charge 3

The location of qo should be at positive x axis, beyond q2(-4q), so has to have a stable charges.

Let the distance of -4q from qo be x

Then from +6q to qo is 0.56+x

The third charge has a charge of q.

Now we need to find Fnet due to charge 3.

Fnet= F13+F23

Let find F13

Let the distance of q1 from charge qo is 0.56+x

Both q1 and q3 are positive, there will be a force of repulsion between them, the F13 will be in the direction of positive x axis

F13=kq1q3/r²

q1= +6q and q3=q

F13=k6qq/(0.56+x)²

F13=k6q²/(0.56+x)²

In vector form

F13=k6q²/(0.56+x)² i

Now let find F23

q2 is negative and q3 is positive, a force or attraction will occur between the two bodies, then the F23 will move in the negative direction of x-axis

Given that, q2=(-4q) and q3=q, r=x

F23=kq1q3/r²

q1= +6q and q3=q

F23=k4qq/x²

F23=k4q²/x²

In vector form

F23=—k4q²/x² i

So, Fnet=F23+F13

Fnet= —k4q²/x²i + k6q²/(0.56+x)² i

Since Fnet=0

Then,

O=—k4q²/x² + k6q²/(0.56+x)²

k4q²/x² = k6q²/(0.56+x)²

Divide through by k2q², then we have

2/x²=3/(0.56+x)²

2(0.56+x)²=3x²

2(0.3136+1.12x+x²)=3x²

0.6272+2.24x+2x²=3x²

3x²-2x²-0.6272-2.24x=0

x²—2.24x—0.6272=0

Using formula method

a =1 b=-2.24 and x=-0.6272

x=-b±sqrt(b²-4ac)/2a

x=2.24±sqrt(-2.24²-4×1×-0.6272)/2×1

x=2.24±sqrt(5.0176+2.5088)/2

x=2.24±sqrt(7.5264)/2

x=(2.24±2.74)/2

Then x=(2.24+2.74)/2

x=2.49

or x=(2.24-2.74)/2

x=-0.25

So, x will be at (0.53+x) from the origin

1. When x =2.49

(0.53+2.49)=3.02m

2. When x =-0.25

This will not be possible because it will be attracted by the charges.

x=-0.25+0.53

x=0.28m

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