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nilankshi
if 2 electrons (a &b) are outside electric field the the potential difference between them will be
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Did you answer your question? The answer is really simple. The electric potential of a negative point charge where there is no electric field (it's the same as 0 at infinity if you do the integral) is the scalar quantity: \[V_E = \frac{1}{4 \pi \epsilon_0}\frac{-q_e}{r}\] If we add another point charge, we really just add the potentials. Since we can define any axis, we can define the horizontal axis so that the other charge is on the point, x=a. Then, the potential is \[V_{E_1+E_2} = \frac{1}{4 \pi \epsilon_0}\frac{-q_e}{r}+ \frac{1}{4 \pi \epsilon_0}\frac{-q_e}{r-a} = \frac{-q_e}{4 \pi \epsilon_0}\Bigg(\frac{1}{r} + \frac{1}{r-a} \Bigg) \] You can reduce that if you would like. To check the result, plug in r = (a/2). Since you are halfway between two equal repulsing charges, you should feel no pressure to move towards either charge and your potential should be zero.
the potential difference between them is zero