Four points a, b, c and d are set at equal distance from the centre of a dipole as shown the figure. The electrostatic potential $V_{a}, V_{b}, V_{c}$ , and $V_{d}$ would satisfy the following relation :

V_{a} > V_{b} > V_{c} > V_{d}

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V_{a} > V_{b} = V_{d} > V_{c}

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V_{a} > V_{c} = V_{b} = V_{d}

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V_{b} = V_{d} > V_{a} > V_{c}

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Solution

The correct option is B $V_{a} > V_{b} = V_{d} > V_{c}$
Here distance between $a$ and $+ q =$ distance between $C$ and $- q = y_{1}$ (say);
distance between $a$ and $- q =$ distance between $C$ and $+ q = y_{2}$
similarly , $d (+ q) = d (- q) = b (- q) = b (+ q) = r$ (say)
Thus , $V_{a} = \frac{k q}{y_{1}} + \frac{- k q}{y_{2}}$
$V_{b} = \frac{k q}{r} + \frac{- k q}{r} = 0$
$V_{c} = \frac{k q}{y_{2}} + \frac{- k q}{y_{1}}$
$V_{d} = \frac{k q}{r} + \frac{- k q}{r} = 0$
Since $y_{2} > y_{1}$ , $V_{a}$ is positive $V_{c}$ is negative.
Thus $V_{a} > V_{b} = V_{d} > V_{c}$

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Four points a, b, c and d are set at equal distance from the centre of a dipole as shown the figure. The electrostatic potential Va,Vb,Vc, and Vd would satisfy the following relation :

Four points a, b, c and d are set at equal distance from the centre of a dipole as shown the figure. The electrostatic potential $V_{a}, V_{b}, V_{c}$ , and $V_{d}$ would satisfy the following relation :