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2 July, 23:23

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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  1. 2 July, 23:48
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    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
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