Question

The electric field strength is 1.70 × 104 N/C inside a parallel-plate capacitor with a 0.800 m spacing. An electron is released from rest at the negative plate. What is the electron's speed when it reaches the positive plate?

Answer 1

The electric field strength between two parallel conducting plates can be determined using the formula E = V/d, where E is the electric field strength, V is the potential difference (voltage) between the plates, and d is the distance between the plates. In this case, the electric field strength is 1.50x10⁶V/m.

Step-by-step explanation:

The electric field strength between two parallel conducting plates can be determined using the formula:

E = V/d

where E is the electric field strength, V is the potential difference (voltage) between the plates, and d is the distance between the plates.

Using the given values, we have:

E = (1.50x104 V) / (1.00 cm)

Converting the distance to meters:

d = 1.00 cm = 0.01 m

Substituting the values:

E = (1.50x104 V) / (0.01 m)

Simplifying the expression:

E = 1.50x106 V/m

Therefore, the electric field strength between the plates is 1.50x106 V/m.