Answer:
E' = Qa/4πε[√(a² + z²)]³
Explanation:
Since the non-conducting plate is symmetric, a small charge element dq generates an electric field dE at a distance R from itself and a distance z above the center of the plate. Since the plate is symmetric, we only have the vertical component of the electric field acting at the center so dE' = dEcosθ where θ is the angle between R and the plate.
So, dE' = dEcosθ = dqcosθ/4πεR²
Let σ represent the surface charge density of the plate. So, for a small elemental area dA, dq = σdA.
Substituting this into dE' we have
dE' = σdAcosθ/4πεR²
Also cosθ = a/R where a is half the length of side of the plate of side length, 2a.
So, dE' = σdAa/4πεR³
Also R² = a² + z²
R = √(a² + z²)
So, dE' = σdAa/4πε[√(a² + z²)]³
Now, dA = dxdy
dE' = σadxdy/4πε[√(a² + z²)]³
So, the total electric field at z is obtain by integrating dE'
E' = ∫dE' = ∫σadxdy/4πε[√(a² + z²)]³ = σa∫dxdy/4πε[√(a² + z²)]³
We integrate dx and dy from -a to a.
So,
E' = σa[2a][2a]/4πε[√(a² + z²)]³
E' = σa³/πε[√(a² + z²)]³
Since the total chare Q = σA where A is the are of the plate. A = (2a)² = 4a²
Q = σA = 4σa²
σ = Q/4a²
substituting σ into E', we have
E' = (Q/4a²)a³/πε[√(a² + z²)]³
E' = Qa/4πε[√(a² + z²)]³
