This is referring to problem 2.25 in Griffiths. Find the potential at a distance 'z' above the midpoint between two equal charges, q, a distance 'd' apart. Compute the electric field in each case.(adsbygoogle = window.adsbygoogle || []).push({});

In the first part the charges are equal. The electric potential is:

[tex]V = \frac{1}{4\pi \epsilon_0} \frac{2q}{{\sqrt{z^2 + (d^2/4)}}}[/tex]

In this case, electric field given by [itex]\vec E = -\nabla V[/itex] is:

[tex]\vec E = \frac{1}{4\pi \epsilon_0} \frac{2qz}{[{z^2 + (d^2/4)}]^{(3/2)}}} \hat z[/tex]

Matches with the result known in problem 2.2(a).

In the 2nd case we consider oppositely charges +q & -q. In this case the electric potential V = 0 which naively() suggests [itex]\vec E = -\nabla V = 0[/itex] which is in contradiction to a previous known result in problem 2.2(b) which suggests:

[tex]\vec E = \frac{1}{4\pi \epsilon_0} \frac{qd}{[{z^2 + (d^2/4)}]^{(3/2)}} \hat z[/tex]

How is this discrepancy accounted for?

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# Homework Help: Finding the electric field in Griffiths

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