Answer :
We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2
F(1) = F (2on1) + F (3on1)
F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N
Since F1 is 7N,
F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)
Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N
F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m
So it's located in -.144m
Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.
Since k = 8.988x10^2, and N=m^2/c^2
F(1) = F (2on1) + F (3on1)
F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N
Since F1 is 7N,
F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)
Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N
F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m
So it's located in -.144m
Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.
Answer:
it is 14.4 cm on -x axis
Explanation:
Force of attraction between q1 and q2 along + x direction is given as
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we have
[tex]F_1 = \frac{(9\times 10^9)(3 \mu C)(5 \mu C)}{0.2^2}[/tex]
[tex]F_1 = 3.375 N[/tex]
Now let say charge q3 is at distance "x" from charge q1
now we know that q3 will attract q1 towards left along - x direction and this force is given as
[tex]F_{net} = 7 N = F_2 - 3.375[/tex]
[tex]F_2 = 10.375 N[/tex]
now we will have
[tex]F_2 = \frac{kq_1q_3}{r^2}[/tex]
[tex]10.375 = \frac{(9\times 10^9)(3 \mu C)(8 \mu C)}{r^2}[/tex]
[tex]r = 0.144 m[/tex]
so it is 14.4 cm on -x axis