A square nonconducting plate, 2a on a side, has a total charge Q uniformly spread over its surface. Calculate the electric field vector E at distance z above the plate's center. Use the electric field at a distance z above the midpoint of a line of charge of length 2L as a shortcut when solving.

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

the answer is c.)

explanation: