Parallel plates: Two flat 4.0 cm × 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite each other. Between the electrodes but not near their edges, the electric field strength is 2.5 × 10⁶ N/C. What is the magnitude of the charge on each electrode? (ε₀ = 8.85 × 10^-12 C²/N ∙ m²)

Question

Quizplus · Accepted Answer

The electric field strength between the plates is given by the formula 
E = V/d, where V is the potential difference between the plates and d is the distance between the plates.

Rearranging this formula, we get: 
V = Ed

Plugging in the given values, we get: 
V = (2.5 x 10^16 N/C) x (2 x 10^-3 m) = 5 x 10^13 V

The potential difference between the plates is equal to the difference in potential between the two plates since they have equal but opposite charges.

So, we have: 
V = Q/ε0A 
where Q is the magnitude of the charge on each plate, ε0 is the permittivity of free space, and A is the area of each plate.

Plugging in the given values, we get: 
5 x 10^13 V = 2Q/(8.85 x 10^-12 C^2/N∙m^2)(0.04 m x 0.04 m) 
Solving for Q, we get: 
Q = 35 x 10^-9 C or 35 nC

Therefore, the magnitude of the charge on each electrode is 35 nC.

Parallel Plates: Two Flat 4