Two oppositely charged but otherwise identical conducting plates of area 2.50 square centimeters are separated by a dielectric 1.80 millimeters thick, with a dielectric constant of k=3.60. the resultant electric field in the dielectric is 1.20×106 volts per meter. part a part complete compute the magnitude of the charge per unit area σ on the conducting plate.

Explanation:

Use the concept of Capacitance of a capacitor with dielectric to solve this problem.

The magnitude of the charge per unit area on the conducting plate is calculated using the expression of charge density.

The magnitude of charge per unit area on the surface of dielectric plate is calculated using the charge density of conducting plate.

The total electric - field energy stored in the capacitor is calculated using the expression for energy stored in the capacitor.

Fundamentals

The expression of charge per unit area on the conducting plate is,

E6 = kEo

Where 6 = magnitude of charge per unit area

E = magnitude of electric feild

Eo= permittivity of free space = 8.854×10^-12

K = dielectric constant

Charge density = 8.854 ×10^-12) × 3.60 × (1.2×10^6)

Magnitude of charge per unit area = 3.82 ×10^-5C^2/Nm^2

volts times amps 6 x 0.5 3w