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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Electrostatics describes forces between charges through Coulomb's law, electric fields and field intensity, then uses electric potential to measure energy per unit charge. Capacitors store charge and electrical energy, with capacitance depending on geometry and dielectric material, not simply on the amount of charge present.
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