Answer:
Step-by-step 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