Two parallel plates that are initially uncharged are separated by 1.7 mm,have only air between them,and each have surface areas of 16 cm².How much charge must be transferred from one plate to the other if 1.9 J of energy are to be stored in the plates? (ε₀ = 8.85 × 10^-12C²/N ∙ m²)

Question

Quizplus · Accepted Answer

The energy stored in a capacitor is given by:
E = (1/2)CV^2
where C is the capacitance and V is the potential difference between the plates.
We can rearrange this equation to solve for C:
C = 2E/V^2
where ε0 is the permittivity of free space.
The capacitance of a parallel-plate capacitor is given by:
C = ε0A/d
where A is the area of the plates and d is the distance between them.
We can combine these two equations to solve for the potential difference between the plates:
V^2 = 2E/(ε0A^2/d)
V = [2Ed/(ε0A^2)]^(1/2)
Finally, we can use the definition of capacitance to solve for the charge Q on each plate:
Q = CV
Substituting the given values and solving for Q, we get:
Q = ε0AV = (ε0A/d)[2E/(ε0A^2/d)]^(1/2) = (2ε0AE/d)^(1/2)
Q = (2 * 8.85x10^-12 C^2/N*m^2 * 0.16 m^2 * 1.9 J)/(0.0017 m)^(1/2) ≈ 5.6 μC

Two Parallel Plates That Are Initially Uncharged Are Separated by 1.7