an infinitely large non conducting plane of uniform charge density σ(sigma) has a circular aperture carved out from it the - Physics - Electric Charges And Fields

Question

an infinitely large non conducting plane of uniform charge
density σ(sigma) has a circular aperture carved out from it
the electric field at a point which is at a distance 'a' from the
centre of the aperture (perpendicular to the plane) isσ/
(2√2Εâ‚€ ) find the radius of the
aperture

Himanshu · Accepted Answer

Electric field intensity at a distance a from the plate is given by: $\frac{σ}{2 ϵ_{0}}$

Electric field intensity at a distance a from the center of the circular aperture and on its axis is given by:

$\frac{σ}{2 ϵ_{0}} (1 - \frac{a}{(a^{2} + R^{2})^{\frac{1}{2}}})$

In the above expression we will replace the positive charge density by negative charge density as this portion of sheet has been removed.

$\frac{- σ}{2 ϵ_{0}} (1 - \frac{a}{(a^{2} + R^{2})^{\frac{1}{2}}})$

Effective electric intensity will be $\frac{σ}{2 ϵ_{0}} + \frac{- σ}{2 ϵ_{0}} (1 - \frac{a}{(a^{2} + R^{2})^{\frac{1}{2}}}) = \frac{σ}{2 ϵ_{0}} (\frac{a}{(a^{2} + R^{2})^{\frac{1}{2}}})$

$\frac{σ}{2 ϵ_{0}} (\frac{a}{(a^{2} + R^{2})^{\frac{1}{2}}}) = \frac{σ}{2 \sqrt{2} ϵ_{0}} \Rightarrow R = a$

Therefore, the radius of the circular aperture will be a

an infinitely large non conducting plane of uniform charge density σ(sigma) has a circular aperture carved out from it.the electric field at a point which is at a distance 'a' from the centre of the aperture (perpendicular to the plane) isσ/ (2√2Εâ‚€.) find the radius of the aperture.