The plates of a parallel-plate capacitor in vacuum are `5.00 mm` apart and `2.00 m^2` in area. A potential difference of `10,000 V` is applied across the capacitor. Compute (a) the capacitance (b) the charge on each plate, and (c) the magnitude of the electric field in the space between them.

The plates of a parallel-plate capacitor in vacuum are `5.00 mm` apart

1.	The plates of a parallel-plate capacitor in vacuum are `5.00 mm` apart and `2.00 m^2` in area. A potential difference of `10,000 V` is applied across the capacitor. Compute (a) the capacitance (b) the charge on each plate, and (c) the magnitude of the electric field in the space between them.
Answer» Correct Answer - `3.54xx10^(9) F ; 3.54xx10^(-6) C ; 2xx10^(5) NC^(-1)` Here, `d = 5xx10^(-3)m, A = 2 m^(2), V = 1000` volt, `C = (in_(0) A)/(d) = (8.85xx10^(-12)xx2)/(5xx10^(-3)) = 3.54xx10^(-9)F` `q = CV = 3.54xx10^(-9)xx1000 = 3.54xx10^(-6) C` `E = (sigma)/(in_(0)) = (Q)/(A in_(0)) = (CV)/(A in_(0)) = (CV)/(in_(0))` `= (3.54xx10^(-9)xx1000)/(2xx8.85xx10^(-12)) = 2xx10^(5) NC^(-1)`

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The plates of a parallel-plate capacitor in vacuum are `5.00 mm` apart and `2.00 m^2` in area. A potential difference of `10,000 V` is applied across the capacitor. Compute (a) the capacitance (b) the charge on each plate, and (c) the magnitude of the electric field in the space between them.

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