The left plate of a parallel plate capacitor carries a positive charge Q, and the right plate carries a negative charge - Q. The magnitude of the electric field between the plates is 100 kV/m. The plates each have an area of 2 x 10-3 m2, and the spacing between the plates is 6 x 10-3 m. There is no dielectric between the plates. What is the charge on the capacitor? Q = The left plate of a parallel plate capacitor carries a positive charge Q, and the right plate carries a negative charge - Q. The magnitude of the electric field between the plates is 100 kV/m. The plates each have an area of 2 x 10-3 m2, and the spacing between the plates is 6 x 10-3 m. There is no dielectric between the plates. What is the magnitude of the potential difference across the capacitor plates? V =

Question

Physics

Dawn

17 October, 21:58

The left plate of a parallel plate capacitor carries a positive charge Q, and the right plate carries a negative charge - Q. The magnitude of the electric field between the plates is 100 kV/m. The plates each have an area of 2 x 10-3 m2, and the spacing between the plates is 6 x 10-3 m. There is no dielectric between the plates. What is the charge on the capacitor? Q = The left plate of a parallel plate capacitor carries a positive charge Q, and the right plate carries a negative charge - Q. The magnitude of the electric field between the plates is 100 kV/m. The plates each have an area of 2 x 10-3 m2, and the spacing between the plates is 6 x 10-3 m. There is no dielectric between the plates. What is the magnitude of the potential difference across the capacitor plates? V =

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Gaige Torres · Answer 1

Capacitance of capacitor

= ε₀ A / d, ε₀ is permittivity of space, A is area of plate, d is distance between plates.

= 8.85 x 10⁻¹² x 2 x 10⁻³ / (6 x 10⁻³)

= 2.95 x 10⁻¹² F

Electric field E = V / d, V is potential difference

V = E x d

= 100 x 10³ x 6 x 10⁻³

= 600 V

Charge on the capacitor

= capacitance x potential difference

= 2.95 x 10⁻¹² x 600

= 17.7 x 10¹⁰ C