Question

Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that are equal in magnitude but opposite in sign. The difference in potential between the plates is 490 V.a. Is the positive or the negative plate at the higher potential?selecti. the positive plateii. the negative plateb. What is the magnitude of the electric field between the plates?____ kV/mc. An electron is released from rest next to the negatively charged surface. Find the work done by the electric field on the electron as the electron moves from the release point to the positive plate. Express your answer in both electron volts and joules.____ eV____ Jd. What is the change in potential energy of the electron when it moves from the release point to the positive plate?_____eVe. What is its kinetic energy when it reaches the positive plate?_____eV

Answer 1

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV