A uniform electric field of E = 5.0×10^6 N/C exists between parallel plates separated by d = 1.0×10^-2 m. What is the potential difference V across the plates?

• A. 5.0×10^4 V
• B. 5.0×10^5 V
• C. 5.0×10^3 V
• D. 5.0×10^2 V

Answer: (A)

In a uniform electric field, the potential difference between two points separated by a distance d changes linearly with distance, so V = E d (the field times the separation). Here E = 5.0×10^6 N/C, which is 5.0×10^6 V/m, and d = 1.0×10^-2 m. Multiply: 5.0×10^6 × 1.0×10^-2 = 5.0×10^4 V. Therefore, the potential difference across the plates is 5.0×10^4 volts (the sign depends on which plate is at higher potential, but the magnitude is 5.0×10^4 V).