A parallel-plate capacitor has an area of 5.00 cm2, and the plates are separated by 2.00 mm with air between them. The capacitor stores a charge of 500 pC. (a) What is the potential difference across the plates of the capacitor? 1 V (b) What is the magnitude of the uniform electric field in the region between the plates?

Area of capacitorA = 5 cm2 = 5*10-4m2distance d= 2 mm = 2*10-3 mpermitivity of the free space ε0 = 8.85*10-12 C2/N.m2charge Q = 500*10-12 CCapacitance of a Capacitor isC = ε0A / d=8.85*10-12*5*10-4m2 / 2*10-3 m=2.21*10-12 Fpotential difference isV= Q/C= 500 *10-12 C/ 2.21*10-12 F = 225.98Vb ) Electric field isE = V / d= 225.98/2*10-3 m=112.99*103 N /C

Question

In: Physics

A parallel-plate capacitor has an area of 5.00 cm2, and the plates are separated by 2.00 mm with...

A parallel-plate capacitor has an area of 5.00 cm2, and the plates are separated by 2.00 mm with air between them. The capacitor stores a charge of 500 pC.
(a) What is the potential difference across the plates of the capacitor?
1 V
(b) What is the magnitude of the uniform electric field in the region between the plates?

Expert Solution

Area of capacitorA = 5 cm² = 5*10^-4m²distance d= 2 mm = 2*10^-3 mpermitivity of the free space ε₀ = 8.85*10^-12 C²/N.m²charge Q = 500*10^-12 CCapacitance of a Capacitor isC = ε₀A / d=8.85*10^-12*5*10^-4m² / 2*10^-3 m=2.21*10^-12 Fpotential difference isV= Q/C= 500 *10^-12 C/ 2.21*10^-12 F = 225.98Vb ) Electric field isE = V / d= 225.98/2*10^-3 m=112.99*10³ N /C

genius_generous answered 2 years ago

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