Find out the P.D. $(V_{B} {- V}_{A})$ for the given system of dipole (r > > a)

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Solution

$V = \frac{k p c o s θ}{r^{2}}$
$V_{B} - V_{A} = \frac{k p}{r^{2}} (cos 45^{\circ} - cos 135^{\circ}) = \frac{\sqrt{2} k p}{r^{2}}$
$p = q a$
$V_{B} - V_{A} = \frac{\sqrt{2}}{4 π ε_{0}} \frac{q a}{r^{2}}$

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Find out the P.D. (VB - VA) for the given system of dipole (r > > a)

V=kpcosθr2VB−VA=kpr2(cos45∘−cos135∘)=√2kpr2p=qaVB−VA=√24πε0qar2

Find out the P.D. $(V_{B} {- V}_{A})$ for the given system of dipole (r > > a)

$V = \frac{k p c o s θ}{r^{2}}$
$V_{B} - V_{A} = \frac{k p}{r^{2}} (cos 45^{\circ} - cos 135^{\circ}) = \frac{\sqrt{2} k p}{r^{2}}$
$p = q a$
$V_{B} - V_{A} = \frac{\sqrt{2}}{4 π ε_{0}} \frac{q a}{r^{2}}$