The Question is from neet-physics electrostatics , and it was created by the author SUBHANI

Question

Two equal and opposite charges $+q$ and $-q$ are separated by a small distance $2a$.  A third charge $+Q$ is placed at a distance $r$ from the midpoint of the dipole along its axial line, where $r >> a$. If the dipole is rotated by $90^\circ$ about its midpoint, the potential energy of the system changes by:

SUBHANI · Accepted Answer

## CORE INFORMATION\n✓ **Answer**: A\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Electric potential energy of a dipole in an electric field\n\t**Related Topics**: Electric dipole, Electric potential, Potential energy\n\t**Prerequisites**: Definition of electric dipole moment, Electric potential due to a dipole, Potential energy of a charge in an electric field\n## SOLUTION PATHWAY\n**Given Information**:\n\t* Two charges, +q and -q, separated by a distance 2a (forming a dipole).\n\t* Dipole moment p = 2aq.\n\t* A third charge +Q is at a distance r from the midpoint of the dipole along its axial line, where r >> a.\n\t* The dipole is rotated by 90 degrees.\n**Method & Steps**:\n\t1.\tCalculate the initial potential energy of the system.\n\t\tReasoning: When the dipole is along the axial line, the potential due to the dipole at the location of charge +Q can be approximated using the far-field approximation.\n\t\tFormula/Rule: The electric potential due to a dipole at a point on the axial line is given by $V = \frac{1}{4\pi\epsilon_0} \frac{p}{r^2}$, where p is the dipole moment and r is the distance from the dipole's center.\n\t2.\tCalculate the potential energy of the charge +Q in the initial dipole field.\n\t\tWorking: The initial potential energy $U_i$ is given by $U_i = QV = \frac{1}{4\pi\epsilon_0} \frac{pQ}{r^2}$.\n\t\tKey Point: We are using the far-field approximation for the potential of the dipole because r >> a.\n\t3.\tCalculate the final potential energy after the dipole is rotated by 90 degrees.\n\t\tReasoning: When the dipole is rotated by 90 degrees, it is now perpendicular to the line connecting the dipole center and the charge +Q. The potential due to the dipole at the location of the charge +Q is now zero.\n\t\tWorking: Therefore, the final potential energy $U_f$ is zero.\n\t4.\tCalculate the change in potential energy.\n\t\tResult: The change in potential energy is $\Delta U = U_f - U_i = 0 - \frac{1}{4\pi\epsilon_0} \frac{pQ}{r^2} = -\frac{pQ}{4\pi\epsilon_0 r^2}$.\n\t\tVerification: The change in potential energy matches with option A.\n ## OPTION ANALYSIS\n[A] ✓:  The change in potential energy is calculated as -pQ/(4πε₀r²), which matches the correct answer.\n[B] :x: This option has the wrong sign. The change in potential energy should be negative as the potential energy decreases after rotation.\n[C] :x: This option has the wrong magnitude. The correct magnitude is pQ/(4πε₀r²).\n[D] :x: The potential energy does change upon rotation. It does not remain zero.\n ## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1.[Error 1]: Forgetting the far-field approximation for the dipole potential. [How to avoid]: Ensure r >> a is satisfied for the approximation to be valid.\n\t\t2.[Error 2]:  Incorrectly determining the potential after rotation. [How to avoid]: When the dipole is perpendicular to the line joining the center and the point, the potential due to the dipole is zero.\n\t**Quick Tips**:\n\t\t1.[Helpful shortcut]: Remember that the potential due to a dipole falls off as 1/r^2.\n\t\t2.[Memory aid]: When calculating potential energy changes, always calculate the initial and final potential energies and then subtract to find the change.

SUBHANI · Answer

The correct Option is "-pQ/(4πε₀r²)"

SUBHANI · Answer

pQ/(4πε₀r²)

SUBHANI · Answer

2pQ/(4πε₀r²)

SUBHANI · Answer

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