The potential difference between the plates of a parallel plate capacitor with the plate separation of 6 cm is 60 V. What is the electric field between the plates of this capacitor?

Question

BlackTom AI · Accepted Answer

Electric field between parallel plates is given by E = V/d for a uniform field, so use that relation directly.

Relevant concepts:
- For an ideal parallel-plate capacitor with plate separation d and potential difference V, the electric field magnitude between the plates is E = V/d (uniform field assumption).
- Always convert lengths to SI units (meters) before calculating field in V/m.

Step-by-step reasoning and calculation:
1. Given V = 60 V and plate separation d = 6 cm.
2. Convert d to meters: 6 cm = 6 × 10^(-2) m = 0.06 m.
3. Apply E = V/d: E = 60 V / 0.06 m.
4. Divide: 60 / 0.06 = 60 × (1/0.06) = 60 × (100/6) = 60 × 16.666... = 1000.
5. Therefore E = 1000 V/m.

Why this answer is correct:
- The formula E = V/d directly gives the electric field for parallel plates; substituting the correctly converted separation yields 1000 V/m. Units check: V/m is the SI unit for electric field, and the numeric division yields 1000, matching that unit.

Why the other options are incorrect:
- 7200 V/m, 2200 V/m, 4400 V/m, and 3600 V/m do not satisfy E = 60/0.06. For example, E = 7200 V/m would require V = E·d = 7200 × 0.06 = 432 V (not 60 V). Similarly, 2200 V/m would correspond to V = 132 V; 4400 V/m to V = 264 V; 3600 V/m to V = 216 V. None match the given 60 V, so they are incorrect.

Conclusion:
The correct electric field between the plates is 1000 V/m, which matches the provided answer.

The potential difference between the plates of a parallel plate capacitor with the plate separation of 6 cm is 60 V. What is the electric field between the plates of this capacitor?单项选择题

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