Problem 5: (8% of Assignment Value) The electric field strength between two parallel conducting plates separated by 4.00 cm is 6.25×10+ V/m. Part (a) ✓ What is the potential difference, in volts, between the parallel conducting plates? V = 2500. V ✓ Correct! Part (b) The electric potential of the lower-potential plate is defined as zero volts. What is the value of the electric potential, in volts, at a point in the gap between the plates that is 2.5 cm from the zero-volt plate? V = || V Grade Summary Deductions 0% 1000f