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A capacitor with a capacitance of 430 pF is connected to a battery with a voltage of 550 V.

Part A) What is the magnitude of the charge on each plate of the capacitor?

Part B) How much energy is stored in the capacitor?

Part C) What is the magnitude of the electric field between the plates if their separation is 0.89 mm?

Respuesta :

Answer:

0.2365 μC, 6.50 x 10⁻⁵ J, 6.28 x 10⁵ V/m

Explanation:

C = 430 pF, V = 550 V, d = 0.89 mm

A) Q = CV = 430 * 550 pC = 236,500 pC = 0.2365 μC

B) U = CV²/2 = 430 x 10⁻¹² x 550²/2 J = 6.50 x 10⁻⁵ J

C) E = V/d = 550/(0.89 x 10⁻³) V/m = 6.28 x 10⁵ V/m

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