On the axis of a uniformly charged disk, what is E_z at z = 0?

On the axis of a uniformly charged disk, what is E_z at z = 0?

• A. E_z = σ/(2 ε0)
• B. E_z = 0
• C. E_z = σ/(2 ε0)
• D. E_z = σ/(ε0)

Answer: (C)

The key idea is the electric field along the axis of a circular disk with uniform surface charge. For a disk of radius R and surface density σ, the axial field at a point a distance z above the center is given by E_z = (σ/(2ε0)) times [1 − z/√(z^2 + R^2)]. This result comes from integrating the contributions of concentric rings and using Coulomb’s law. Now, evaluate at z → 0+ (just above the disk). The term z/√(z^2 + R^2) tends to 0, so E_z → σ/(2ε0). Therefore, just above the disk, the axial field has magnitude σ/(2ε0) and points away from the disk for positive σ. If you consider the exact plane, the field is not defined there, but the limit from above gives the standard value.

The key idea is the electric field along the axis of a circular disk with uniform surface charge. For a disk of radius R and surface density σ, the axial field at a point a distance z above the center is given by E_z = (σ/(2ε0)) times [1 − z/√(z^2 + R^2)]. This result comes from integrating the contributions of concentric rings and using Coulomb’s law.

Now, evaluate at z → 0+ (just above the disk). The term z/√(z^2 + R^2) tends to 0, so E_z → σ/(2ε0). Therefore, just above the disk, the axial field has magnitude σ/(2ε0) and points away from the disk for positive σ. If you consider the exact plane, the field is not defined there, but the limit from above gives the standard value.