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 Compute the magnitude of the charge per unit area σ on the conducting plate. Express your answer in coulombs per square meter to three significant figures.

Question

Physics

Mara Chen

22 May, 03:24

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 Compute the magnitude of the charge per unit area σ on the conducting plate. Express your answer in coulombs per square meter to three significant figures.

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Rayan Stuart · Answer 1

Answer: 38.2 μC

Explanation: In order to solve this problem we have to use the relationship for a two plate capacitor with a dielectric so:

C = Q/V = we also know that for two paralel plates C=εo*k*A/d and V=E/d

where k is the dielectric constant, A plate area, V is potential difference; E electric field and d the separation between the plates.

reorganizing we have:

Q/A=σ = E*k/εo = 1.2 * 10^6*3.6/8.85 * 10^-12=38.2μC