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Three point charges are arranged along the x-axis. Charge q1 = +3.00 uC is at the origin, and charge q2= -5.00 uC is at x= 0.200 . Charge q3= -8.00 uC.

Where is q3 located if the net force on q1 is 7.00 N in the -x direction? ...?

Verified Answer :

MissPhiladelphia

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. 

Answer:

it is 14.4 cm on -x axis

Explanation:

Force of attraction between q1 and q2 along + x direction is given as

\(F = \frac{kq_1q_2}{r^2}\)

here we have

\(F_1 = \frac{(9\times 10^9)(3 \mu C)(5 \mu C)}{0.2^2}\)

\(F_1 = 3.375 N\)

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

\(F_{net} = 7 N = F_2 - 3.375\)

\(F_2 = 10.375 N\)

now we will have

\(F_2 = \frac{kq_1q_3}{r^2}\)

\(10.375 = \frac{(9\times 10^9)(3 \mu C)(8 \mu C)}{r^2}\)

\(r = 0.144 m\)

so it is 14.4 cm on -x axis

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