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Question #42947

Difficulty: 8/10 | 🟢 Free

The Question

In a certain region of space, the electric potential is given by V(x,y,z)=α(x2+y2−2z2)V(x,y,z) = \alpha (x^2 + y^2 - 2z^2), where α\alpha is a positive constant with units of V⋅m−2\text{V}\cdot\text{m}^{-2}, and x,y,zx, y, z are Cartesian coordinates in meters. What is the total electrostatic energy stored in the electric field within a spherical volume of radius RR centered at the origin?

Options

  • A.

    16πϵ0α2R55\frac{16\pi \epsilon_0 \alpha^2 R^5}{5}

  • B.

    00

  • C.

    8πϵ0α2R55\frac{8\pi \epsilon_0 \alpha^2 R^5}{5}

  • D.

    32πϵ0α2R55\frac{32\pi \epsilon_0 \alpha^2 R^5}{5}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Variable_Constraint

Develop pure symbolic logic skills by mastering algebraic constraints without relying on numerical data.

Concepts Behind This Question

Potential Gradient Direction|Energy Density Locality|Self-Energy of Charged Sphere

Why This Question Is Tricky

Common Mistakes

This question tests the fundamental definition of electrostatic energy stored in an electric field.

A common trap for advanced students is to calculate the volume charge density using Poisson's equation (ρ=−ϵ0∇2V\rho = -\epsilon_0 \nabla^2 V).

For this specific quadrupole potential, ∇2V=2α+2α−4α=0\nabla^2 V = 2\alpha + 2\alpha - 4\alpha = 0, meaning the region is completely charge-free (ρ=0\rho = 0).

Students relying on the formula U=12∫ρVdτU = \frac{1}{2} \int \rho V d\tau will incorrectly conclude the stored energy is zero (Option B).

However, that formula represents the total energy of a complete charge distribution over all space, not the energy localized within a specific volume.

To find the energy stored in a specific region, one must strictly integrate the local energy density u=12ϵ0E2u = \frac{1}{2}\epsilon_0 E^2 over that volume.

Furthermore, the non-spherical nature of the potential gradient forces students to perform a full 3D volume integral in spherical coordinates, preventing algebraic shortcuts and penalizing those who incorrectly assume ∣E⃗∣2∝r2|\vec{E}|^2 \propto r^2 symmetrically (Option C).

Quick Facts About This Question

Question Information

SubjectPhysics
ChapterElectrostatic Potential and Capacitance - Class 12
ConceptPotential Gradient Direction
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree