ELECTROSTATICS cYU 4

Solution: Electric Field of Half-Planes

Problem 4: Parallel Charged Half-Planes

Question: Two parallel half-planes are uniformly charged with surface charge densities \(+\sigma\) and \(-\sigma\). Find the magnitude and direction of the electric field due to these half-planes at a point P located at a height \(h\) above the edge of the positively charged half-plane. Distance \(d\) between the half-planes is much smaller than \(h\) (\(d \ll h\)).

Conceptual Analysis

Since the distance between the plates \(d\) is much smaller than the observation distance \(h\) (\(d \ll h\)), we can approximate the system. The two semi-infinite sheets act like a long “strip” or a line charge at large distances.

The system is effectively a linear charge distribution located at the edge.

1. Calculate Linear Charge Density (\(\lambda\))

Consider a small cross-sectional length. The effective linear charge density \(\lambda\) is the surface charge density \(\sigma\) multiplied by the width of the separation \(d\).

$$ \lambda = \sigma \cdot d $$

2. Electric Field Calculation

We treat the setup as an infinite line charge running along the edge axis. The formula for the electric field of an infinite line charge at a distance \(r\) is:

$$ E = \frac{\lambda}{2\pi \epsilon_0 r} $$

Here, the distance \(r\) is given as \(h\). Substituting \(\lambda = \sigma d\):

$$ E = \frac{\sigma d}{2\pi \epsilon_0 h} $$

3. Direction

The field lines for a dipole-like strip originate from the positive sheet and terminate on the negative sheet. At point P (directly above the edge), the field vector points horizontally, parallel to the planes, directed away from the positive region towards the negative region’s side (perpendicular to the edge line).

Final Answer

$$ E = \frac{\sigma d}{2\pi \epsilon_0 h} $$

Direction: Parallel to the planes, perpendicular to the edge (from the \(+\sigma\) side plane towards the \(-\sigma\) side).