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A dipole of dipole moment vecp=phati li...

A dipole of dipole moment `vecp=phati` lies along the x -axis in a non-uniform electric field `vecE=(c)/(x)hati` . The force acting on the dipole is

A

zero

B

`-(pc)/(x^(2))hati`

C

`(pc)/(x^(2))hati`

D

`-(pc)/(2x^(2))hati`

Text Solution

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The correct Answer is:
To solve the problem of finding the force acting on a dipole in a non-uniform electric field, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the Dipole and Electric Field**: - A dipole consists of two equal and opposite charges separated by a small distance. The dipole moment \(\vec{p}\) is given as \(\vec{p} = p \hat{i}\). - The electric field is given as \(\vec{E} = \frac{c}{x} \hat{i}\), which is a non-uniform electric field that varies with \(x\). 2. **Determine the Forces on the Charges**: - Let the two charges of the dipole be \(+q\) and \(-q\), separated by a small distance \(A\). - The force on the positive charge \(+q\) in the electric field \(\vec{E}\) is: \[ F_+ = +q \cdot \vec{E} = +q \cdot \frac{c}{x} \hat{i} \] - The force on the negative charge \(-q\) is: \[ F_- = -q \cdot \vec{E} = -q \cdot \frac{c}{x} \hat{i} \] 3. **Calculate the Change in Electric Field**: - Since the electric field is non-uniform, we need to consider the change in the electric field over the distance \(A\). - The change in electric field \(dE\) can be calculated using the derivative of \(\vec{E}\): \[ \frac{dE}{dx} = -\frac{c}{x^2} \] - For a small displacement \(dx = A\), the change in electric field \(dE\) is: \[ dE = \frac{dE}{dx} \cdot A = -\frac{c}{x^2} \cdot A \] 4. **Net Force on the Dipole**: - The net force \(F_{net}\) acting on the dipole can be expressed as the difference between the forces on the two charges: \[ F_{net} = F_+ - F_- = q \cdot \frac{c}{x} - (-q \cdot \frac{c}{x + A}) \] - Approximating for small \(A\), we can express the net force as: \[ F_{net} = q \cdot dE = q \cdot \left(-\frac{c}{x^2} A\right) \] - Since \(p = qA\), we can rewrite the force as: \[ F_{net} = -\frac{p \cdot c}{x^2} \] 5. **Direction of the Force**: - The direction of the force is determined by the sign of the change in electric field. Since the electric field decreases with increasing \(x\), the positive charge experiences a smaller force than the negative charge. Therefore, the net force will be directed in the negative \(x\) direction, or \(-\hat{i}\). ### Final Result: Thus, the force acting on the dipole is: \[ \vec{F} = -\frac{p \cdot c}{x^2} \hat{i} \]
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