A dipole of dipole moment 'p' is placed in non-uniform electric field along x-axis. Electric field is increasing at the rate of $1 V m^{- 1}$ . The force on dipole is:

0

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2 p

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p / 2

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p

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Solution

The correct option is B $p$
Force is given as the negative gradient of potential energy. $F = \frac{- d U}{d x}$
Potential energy for an electric dipole is $U = - \to p . \to E$
So, we get the force as, $F = p \frac{d E}{d x} = p$
As
$\frac{d E}{d x} = 1$

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A dipole of dipole moment 'p' is placed in non-uniform electric field along x-axis. Electric field is increasing at the rate of 1 Vm−1. The force on dipole is:

The correct option is B pForce is given as the negative gradient of potential energy. F=−dUdxPotential energy for an electric dipole is U=−→p.→ESo, we get the force as, F=pdEdx=pAs dEdx=1

A dipole of dipole moment 'p' is placed in non-uniform electric field along x-axis. Electric field is increasing at the rate of $1 V m^{- 1}$ . The force on dipole is:

The correct option is B $p$
Force is given as the negative gradient of potential energy. $F = \frac{- d U}{d x}$
Potential energy for an electric dipole is $U = - \to p . \to E$
So, we get the force as, $F = p \frac{d E}{d x} = p$
As
$\frac{d E}{d x} = 1$