Two large, parallel, conducting plates are 15 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. An electrostatic force of 4.8 × 10-15 N acts on an electron placed anywhere between the two plates. (Neglect fringing.) (a) Find the electric field at the position of the electron. (b) What is the potential difference in volts between the plates?

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Asked Jun 20, 2019
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Two large, parallel, conducting plates are 15 cm apart and have charges of equal magnitude and opposite sign on their facing surfaces. An electrostatic force of 4.8 × 10-15 N acts on an electron placed anywhere between the two plates. (Neglect fringing.) (a) Find the electric field at the position of the electron. (b) What is the potential difference in volts between the plates?

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Expert Answer

Step 1

Given

Distance between the plates d = 15 cm = 0.15 m

Electrostatic force on electron F = 4.8 x 10-15 N

Known value

Charge of an electron q = 1.6 x 10-19 C

Solution

A) Finding the electric field

Electric Field=Electric Force
Charge
F
E =
4.8 x1015
E =
1.6x101
E 3x10 N
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Electric Field=Electric Force Charge F E = 4.8 x1015 E = 1.6x101 E 3x10 N

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Step 2

B) Finding Potential Differ...

Potential difference Electric field . Distance
AV = Ed
AV 3x10 x0.15
AV 4.5x10
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Potential difference Electric field . Distance AV = Ed AV 3x10 x0.15 AV 4.5x10

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