A 10 nanofarad (10 x 10^-9 F) parallel plate capacitor holds a charge of magnitude 50 µC on each plate.If the plates are separated by a distance of 0.885mm, what is the area of each plate?
Correct answer: A. 1.0 m2
- A. 1.0 m2
- B. 3.0 m2
- C. 5.5 m2
- D. 7.5 m2
Explanation
To find the area of each plate, we use the formula for the capacitance of a parallel plate capacitor: C = ε0 * (A/d), where C is the capacitance, ε0 is the permittivity of free space (approximately 8.85 x 10-12 F/m), A is the area of the plates, and d is the distance between the plates. Given C = 10 x 10-9 F and d = 0.885 x 10-3 m, we rearrange the formula to solve for A: A = (C * d) / ε0. Plugging in the values gives A = (10 x 10-9 F * 0.885 x 10-3 m) / 8.85 x 10-12 F/m = 1.0 m2. Thus, Option A is correct. The other options provide areas that are either too large or too small when calculated properly with the given values.
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