A parallel-plate capacitor has plate area of 1.0 cm by 1.0 cm and plate separation of 3.0 mm. The internal electric field is 1.5×10^5 V/m. What is the potential difference across the plates?

• A. 300 V
• B. 450 V
• C. 600 V
• D. 150 V

Answer: (B)

Between the plates, the potential difference V is the product of the uniform field and the separation: V = E d. The separation is 3.0 mm = 0.003 m, and the field is 1.5×10^5 V/m, so V = (1.5×10^5)(0.003) = 450 V. The plate area doesn’t affect V here (it would matter for capacitance and charge, via C = ε0 A / d and Q = C V). So the potential difference is 450 V.