Net Electric Field Equals Zero

[AFRaven]

Homework Statement

Two charges, -29.7$$\mu$$C and +5.5$$\mu$$C, are fixed in place and separated by 4m
(a) At what spot along the line through the charges is the net electric field zero? Locate this spot relative to the position of the positive charge
(b) What would the magnitude of the force on a third charge +48.7$$\mu$$C placed at this spot

Homework Equations

E=kq/d$$^{2}$$

The Attempt at a Solution

......+5.5microC...-29.7microC
o_________________o______________o_________
|---------d---------|------4m-------|

After drawing out the problem as shown above, I found out that the net electric field would be zero to the left of the positive charge (-x direction). I then set up my equation as follows:

E= k(5.5*10^-6)/d$$^{2}$$ - k(2.97*10^-7)/(4+d)$$^{2}$$

I graphed this formula and came up with -5.2109 or -3.24575. I entered both numbers into the system as positive and negative and came up with the wrong answer every time. Also, once I solve for "d", how do I go about finding the magnitude of force on the third charge?

 

[Carid]

29.7microC = 2.97*10^-5 C

 

[thomate1]

I would like to provide the following response to the above content:

The net electric field is determined by the sum of all the electric fields from the individual charges. When the net electric field is zero, it means that the electric field from one charge is equal in magnitude and opposite in direction to the electric field from the other charge. In this case, the net electric field is zero to the left of the positive charge, as correctly identified.

To find the exact spot where the net electric field is zero, we can use the formula E = kq/d^2, where k is the Coulomb's constant, q is the charge, and d is the distance between the charges. We can set up an equation with the electric fields from the two charges and solve for d. However, it is important to note that the distance d in this formula is the distance between the point where the net electric field is zero and the positive charge, not the distance between the two charges. Therefore, the equation should be:

E = kq/d^2 - kq/(4-d)^2

Solving for d, we get d = 1.6m. This means that the spot where the net electric field is zero is 1.6m to the left of the positive charge.

To find the magnitude of the force on a third charge placed at this spot, we can use the formula F = qE, where F is the force, q is the charge, and E is the electric field at that spot. Plugging in the values, we get F = (48.7*10^-6)(5.5*10^6)/1.6^2 = 0.97N. Therefore, the magnitude of the force on the third charge would be 0.97N.