A positive point charge q = +2.50 nC is located at x = 1.20 m and a negative cha

Question

A positive point charge q = +2.50 nC is located at x = 1.20 m and a negative charge of -2q = -5.00 nC is located at the origin as shown in Figure P16.18. (a) Sketch the electric potential verses x for points on the x-axis in the range -1.50 m < x < 1.50 m. (b) Find a symbolic expression for the potential on the x-axis at an arbitrary point P between the two charges. The expression should be in terms of x, q and d = 1.20 m, the distance between the two charges. (c) Find the electric potential at x = 0.600 m, i.e. at x = d/2. (d) Find the value of x at the point along the x-axis between the two charges where the electric potential is zero.

Explanation / Answer

You just have to add two vectors. Vector1 is F1 force. Vector2 is F2 force. F1x - x component of F1; F1y - y component of F1. F2x, F2y - the same of F2 You can see that F1x = -F2x and they are opposite directions. So, to find the resulting force F we just have to add F1y and F2y. F = F1y+F2y. F2y = F1y it's clear. F = 2*F1y F1y = k*q*q/r^2 * cos (atan (d/h)), where r^2 = h^2+d^2. So, F = 2 * k * q*q/(h^2+d^2) * cos (atan(d/h)). (1) To answer the second part of your question we need to find derivative F'(h). Sorry, it would take time. To find optimal h you need to solve F'(h) = 0. Good luck! Well, I've tried. There is a lot to write. The answer is h = d/sqrt(2).

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