Question

HELP ASAPP PLEASEE

The electric field intensity between two charged plates is 4.1 x 10 N/C. The plates are 0.01 m apart. What is the potential difference between the plates in volts?

Answer 1

Answer: It can be moved 0.01 m toward the negative plate.

Explanation: I just took the test.

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Answer 2

To calculate the Potential Difference between the plates in an Uniform Electric Field we have the formula:

`u = e * d`
U = Potential difference = ?
E = Electric Field = 4.1 x 10 N/C
D = Distance = 0.01m = 1 x 10^-2 m


`u = (4.1 * 10) * (1 * 10^( - 2)) \\ u = 41 * 10^( - 2) \: volts \: or \: \: 0.41 \: volts`