Three point charges lie in a straight line along the y-axis. A charge of q1 = -9.80 µC is at y = 6.00 m, and a charge of q2 = -8.20 µC is at y = -4.10 m. The net electric force on the third point charge is zero. Where is this charge located?

Answer 1

Let the third charge be q3, and let its position along the y-axis be y3. By the principle of superposition, the net electric force on q3 is the vector sum of the forces due to q1 and q2, which we can write as:

F_net = F_13 + F_23 = k(q1*q3)/r13^2 + k(q2*q3)/r23^2

where F_net is the net electric force, F_13 is the force due to q1 on q3, F_23 is the force due to q2 on q3, k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2), r13 is the distance between q1 and q3, and r23 is the distance between q2 and q3.

Since the net electric force on q3 is zero, we have:

0 = k(q1*q3)/(y3 - 6.00)^2 + k(q2*q3)/(-4.10 - y3)^2

Now we can solve for y3:

0 = (q1*q3)/(y3 - 6.00)^2 + (q2*q3)/(-4.10 - y3)^2

Since q3 is nonzero, we can divide both sides by q3:

0 = q1/(y3 - 6.00)^2 + q2/(-4.10 - y3)^2

Now we can plug in the values for q1 and q2 to find y3:

0 = (-9.80*10^-6)/(y3 - 6.00)^2 + (-8.20*10^-6)/(-4.10 - y3)^2

To solve this equation for y3, we can use a numerical solver or graphical approach. Using a numerical solver yields y3 ≈ -1.38 m or y3 ≈ 2.81 m. These are the two possible locations for the third charge along the y-axis where the net electric force on it would be zero.