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The plates of a parallel-plate capacitor are 2.50 mm
Chapter 24, Problem 1E(choose chapter or problem)
The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00 X 106 V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?
Questions & Answers
QUESTION:
The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00 X 106 V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?
ANSWER:Solution 1E The relation between potential difference and the electric field between the plates is given by V = Ed. Given in the question, electric field E = 4.00 × 10 V /m 3 Distance between the plates d = 2.50 mm = 2.50 × 10 m (a) Therefore, the potential difference V = 4.00 × 10 V /m × 2.50 × 10 3m = 10 × 10 V = 10.0 kV So, the potential