Animated Solution for Physics - Electrostatics: An electric dipole with dipole moment p0=2p0(i^+j^) is held fixed at the origin O in the presence of an uniform electric field of magnitude E0. If the potential is constant on a circle of radius R centered at the origin as shown in figure, then the correct statement(s) is/are:
(ε0 is permittivity of free space, R>> dipole size)
Select Answer:
* Multiple Correct
Visualized Solution
VisualizingtheSetup
Dipole p0=2p0(i^+j^) is at 45∘.
Circle of radius R is an equipotential surface.
ConditionforEquipotentialSurface
Vtotal=Vdipole+Vuniform=constant
Potential must be independent of angle θ.
PotentialExpressions
Vdipole=4πε01R2p0cos(θ−45∘)
Let E0 be at angle α.
Vuniform=−E0Rcos(θ−α)
MakingPotentialConstant
Vtotal=4πε0R2p0cos(θ−45∘)−E0Rcos(θ−α)=C
To cancel θ dependence: α=45∘
CalculatingRadiusR
Coefficients must be equal:
4πε0R2p0=E0R
R3=4πε0E0p0⟹R=(4πε0E0p0)1/3
ElectricFieldatPointA
Point A is on the axis of the dipole.
Ed,A=4πε0R32p0p^
Since E0=4πε0R3p0, Ed,A=2E0p^
TotalFieldatA
EA=Ed,A+E0
EA=2E0p^+E0p^=3E0p^
EA=3E02(i^+j^) (Option C is incorrect)
ElectricFieldatPointB
Point B is on the equatorial line of the dipole.
Ed,B=−4πε0R3p0p^
Ed,B=−E0p^
TotalFieldatB
EB=Ed,B+E0
EB=−E0p^+E0p^=0
(Option D is correct)
Conclusion
EA=3E0 and EB=0
Magnitudes are not equal (Option B is incorrect).
Correct Options: (A) and (D)
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The Sigma Insight: Electric Dipole
Solution Diagram
Imagine a tiny electric dipole sitting at the origin, pointing diagonally at a 45-degree angle. Now, bathe this entire setup in a uniform electric field. The problem tells us something magical happens: a perfect circle of radius R drawn around the origin becomes an equipotential surface. This single geometric fact is the key that unlocks the entire problem.
Analyzing the Setup
What does it mean for a circle to be an equipotential surface? It means that if you walk along the perimeter of this circle, you will not experience any change in electric potential
The total potential V is a constant value, completely independent of your angular position θ.
The total potential at any point in space is simply the scalar sum of the potentials created by the individual sources. Here, we have two sources: the dipole and the uniform electric field.
The Master Equation
Let's write down the potential for each
The potential due to a dipole at a distance R and angle θ from its axis is given by:
Vdipole=4πε01R2p0cos(θ−45∘)
Notice that we use (θ−45∘) because the dipole itself is tilted at 45∘ to the x-axis.
Now, what about the uniform electric field E0? Let's assume it points in some arbitrary direction α. The potential of a uniform field is V=−E⋅r. In polar coordinates, this becomes:
Vuniform=−E0Rcos(θ−α)
Adding them together gives us our master equation for the total potential on the circle:
Vtotal=4πε0R2p0cos(θ−45∘)−E0Rcos(θ−α)
The Grand Cancellation
Here is where the physics gets beautiful
We know Vtotal must be a constant, meaning it cannot depend on θ. But our equation is full of θs! The only way this equation can be independent of θ is if the two cosine terms perfectly cancel each other out for every possible value of θ.
For this perfect cancellation to occur, two conditions must be met simultaneously. First, the angular dependence must be identical, which forces the uniform field to align perfectly with the dipole:
α=45∘
Second, the coefficients of the cosine terms must be equal in magnitude:
4πε0R2p0=E0R
Solving this elegant little equation for R gives us:
R=(4πε0E0p0)1/3
This perfectly matches Option (A)!
Final Calculation
Fields at A and B
Now that we know the uniform field E0 points in the exact same direction as the dipole moment p0, we can easily find the total electric field at specific points using the principle of superposition.
Point A lies directly on the axis of the dipole. The electric field of a dipole on its axis points in the same direction as the dipole moment and has a magnitude of 2kp0/R3.
Substituting our finding that E0=kp0/R3, the dipole's field at A is simply 2E0.
Adding the uniform field E0 (which points in the same direction), the total field at A is:
EA=2E0p^+E0p^=3E0p^
Since p^=2i^+j^, the field is 3E02i^+j^. This makes Option (C) incorrect.
Point B lies on the equatorial line of the dipole. The electric field of a dipole on its equator points opposite to the dipole moment and has a magnitude of kp0/R3, which is exactly E0.
So, the dipole's field at B is −E0p^.
Adding the uniform field E0p^, we get a spectacular cancellation:
EB=−E0p^+E0p^=0
This confirms Option (D) is absolutely correct.
Because the field is 3E0 at A and 0 at B, the magnitude is clearly not the same everywhere, rendering Option (B) incorrect. The physics of equipotential surfaces has guided us flawlessly to the solution!
