# Question about Coulomb's Law and attraction of charges in uneven field

I am trying to find the formula using Coulomb's Law for below drawing where instead of having two charges in the same charged field, the field strength is different between the two. I understand Coulomb's law as

$$F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{d^2}$$

is there some way to correct for the additional uneven or additional attraction based on the field strength? I tried to make a drawing to show what I mean. Thanks for any suggestions.

From the picture I'm just going to assume that the problem takes place in the plane. Moreover, it looks like potential is linearly increasing with horizontal distance $$x$$, so its derivative is constant. So we can write the external potential from

$$\frac{\partial V(x,y)}{\partial x} = V_0 \implies V(x,y) = V_0 x.$$

where I have ignored the arbitrary constant of integration.

Your total potential at a given point $$(x,y)$$ should then be

$$V(x,y) = V_{1} + V_{2} + V_0x$$

where $$V_i = \frac{1}{4\pi\epsilon_0} \frac{q_i}{\sqrt{(x-x_i)^2 + (y-y_i)^2}}$$. You can now easily calculate the force on $$q_1, q_2$$.

• Don't be a freeloader now. If it helped you either upvote or accept the answer if it answered your question Commented Mar 21, 2019 at 21:13