electric field derived from potential?

1. The charge Q at (0,a) will give a potnetial at point (x,0) given by
V(x) = k.Q/sqrt(a^2+x^2)
where k is a constant(=9x10^9 N.m^2/C^2)
[sqrt stands for square-root]
By symmetry, the charge at (0,-a) also gives the same potnetial at (x,0)
thus the total potnetial V(x) = 2kQ/sqrt(a^2+x^2)
(b) Using E=-dV(x)/dx, you can calculate the required electric field E in terms of x
E(x) = 2kQx/(a^2+x^2)^(3/2)
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2. First calculate the charge separation along the x, y and z axis.
x-axis: 2-(-1) units = 3 units
y-axis: 3-4 units = -1 units
z-axis: -5-2 units = -7 units
The separation between the two charges, s = sqrt[3^2+(-1)^2+(-7)^2] units
The potential energy = k[Q1][Q2]/s
where k is defined as before. Substitute the value of Q1, Q2 and s into the equations to calculate the answer.