Question

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 52000 m/s. What is the potential difference between the plates of the capacitor?

1) Cannot be determined
2) 52000 V
3) 104000 V
4) 26000 V

Answer 1

To find the potential difference between the plates of a capacitor, we can use the equation V = Q/C, where V is the potential difference, Q is the charge on the plates, and C is the capacitance of the plates. In this case, we are given the charge on the plates (0.020 µC) and the potential difference (250 V), so we can rearrange the equation to solve for the capacitance. Substituting the given values, the capacitance of the plates is 0.00008 F.

Step-by-step explanation:

To find the potential difference between the plates of a capacitor, we can use the equation:
V = Q/C

Where V is the potential difference, Q is the charge on the plates, and C is the capacitance of the plates.

In this case, we are given the charge on the plates (0.020 µC) and the potential difference (250 V), so we can rearrange the equation to solve for the capacitance:

C = Q/V

Substituting the given values:

C = (0.020 µC) / (250 V) = 0.00008 F

Therefore, the capacitance of the plates is 0.00008 F.