Where a proton would experience zero net force

Question

Consider two charges, where one (+3.80 nC) is at the origin and the other (-14.4 nC) is at the position x = 4.51 cm.

Find the x-coordinate where a proton would experience zero net force.

I am completely stuck on this question

Answer 1

Because you have a + charge and a larger - charge, you know that for the net force on the proton to be zero, you need the proton to be placed to the left of the + charge. Between the charges is wrong because both charges will apply forces to the right, and to the right of the - charge cannot result in a balance because the attraction of the larger and nearer charge cannot be balanced by the repulsion from the farther and weaker charge.

Lets place the x to the left of the positive charge (This will be position -x).

Magnitude of repulsion to the left from positive charge is k_e(3.8 x 10^-9C)q_e/x² (q_e is charge on proton/electron)

Magnitude of attraction to the rights from negative charge is k_e(14.4 x 10^-9C)q_e/(.0451+x)²

Equate these forces and cancel factorable quantities: 3.8/x² = 14.4/(4.51+x)² (now that a proportionality is involved, units just need to match (They don't have to be MKS - numerator is in nC and denominator is in cm²)

Now solve this for x by cross-multiplying and grouping terms. I'll leave that to you.

Please consider a tutor. Take care.