If a dipole of charge 2 …
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Sia ? 6 years, 8 months ago
{tex}\phi = \frac{q}{\epsilon}{/tex}
where, {tex}\phi{/tex} is electric flux,
q is the net charge
Since dipole has +q and -q charges
Therefore, the net charge is 0.
Therefore, {tex}\phi = \frac{0}{\epsilon}{/tex}=0
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