Question

The plates of a parallel-plate capacitor are 3.00 mm apart, and each carries a charge of magnitude 85.0nC. The plates are in vacuum. The electric field between the plates has a magnitude of 5.00×10⁶ V/m. Part A What is the potential difference between the plates?

Answer 1

The potential difference between the plates of a parallel-plate capacitor is calculated by multiplying the electric field by the distance between the plates, resulting in a potential difference of 15000 V (or 15 kV).

Step-by-step explanation:

The potential difference between the plates of a parallel-plate capacitor can be calculated using the relation between the electric field (E) and the potential difference (V), which is V = E * d, where d is the distance between the plates. Given that the electric field (E) is 5.00×10⁶ V/m and the separation between the plates is 3.00 mm (or 0.003 m), we can calculate the potential difference as follows: V = E * d = (5.00×10⁶ V/m) * (0.003 m) = 15000 V. Therefore, the potential difference between the plates is 15000 V (or 15 kV).