# Problem: The electric field strength between two parallel conducting plates separated by 4.00 cm is 7.50x104 V/m. (a) What is the potential difference between the plate?(b) The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that plate (and 3.00 cm from the other)?

###### FREE Expert Solution

The voltage between the parallel conducting plates is given by:

$\overline{){\mathbf{V}}{\mathbf{=}}{\mathbf{E}}{\mathbf{d}}}$

83% (142 ratings) ###### Problem Details

The electric field strength between two parallel conducting plates separated by 4.00 cm is 7.50x104 V/m.

(a) What is the potential difference between the plate?

(b) The plate with the lowest potential is taken to be at zero volts. What is the potential 1.00 cm from that plate (and 3.00 cm from the other)?

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