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elena-s [515]
3 years ago
7

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
Physics
1 answer:
elena-s [515]3 years ago
7 0

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

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A car can travel 100 m in 5 seconds

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(1) A positive charge +3 C is separated from another positive charge of +5 C by a distance of 7m. What is the magnitude of the e
Aneli [31]

1. The magnitude of the electric force between the two charges is 2.8×10⁹ N (Option B)

2. The net charge on the molecule is -8×10⁻¹⁹ C (Option D)

3. The magnitude of the force between the charges is 16000 N (Option C)

4. The correct statement is: A neutral object has equal numbers of protons and electrons. (Option C)

<h3>1. How to determine the force</h3>
  • Charge 1 (q₁) = +3 C
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  • Electric constant (K) = 9×10⁹ Nm²/C²
  • Distance apart (r) = 7 m
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F = Kq₁q₂ / r²

F = (9×10⁹ × 3 × 5) / (7)²

F = 2.8×10⁹ N

<h3>2. How to determine the net charge on the molecule</h3>
  • Electron = 223 electrons
  • Proton = 218 protons
  • Net Charge =?

Charge = Proton - Electron

Charge = 218 - 223

Charge = -5 electrons

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1 electron = 1.6×10⁻¹⁹ C

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Net Charge = -8×10⁻¹⁹ C

<h3>3. How to determine the force</h3>
  • Charge 1 (q₁) = 2×10⁻⁴ C
  • Charge 2 (q₂) = 8×10⁻⁴ C
  • Electric constant (K) = 9×10⁹ Nm²/C²
  • Distance apart (r) = 0.3 m
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F = Kq₁q₂ / r²

F = (9×10⁹ × 2×10⁻⁴ × 8×10⁻⁴) / (0.3)²

F = 16000 N

<h3>4. What is a neutral object?</h3>

A neutral object is an object having equal numbers of protons and electrons. For example, an object with 4 protons and 4 electrons is said to be neutral as illustrated below

  • Electron = 4 electrons
  • Proton = 4 protons
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Charge = Proton - Electron

Charge = 4 - 4

Charge = 0 (neutral)

Thus, the correct statement about neutral object, given in the question is: A neutral object has equal numbers of protons and electrons (Option C)

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6 0
2 years ago
A park ranger driving on a back country road suddenly sees a deer in his headlights 20
olya-2409 [2.1K]

Answer:

17.1

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The distance ahead, of the deer when it is sighted by the park ranger, d = 20 m

The initial speed with which the ranger was driving, u = 11.4 m/s

The acceleration rate with which the ranger slows down, a = (-)3.80 m/s² (For a vehicle slowing down, the acceleration is negative)

The distance required for the ranger to come to rest, s = Required

The kinematic equation of motion that can be used to find the distance the ranger's vehicle travels before coming to rest (the distance 's'), is given as follows;

v² = u² + 2·a·s

∴ s = (v² - u²)/(2·a)

Where;

v = The final velocity = 0 m/s (the vehicle comes to rest (stops))

Plugging in the values for 'v', 'u', and 'a', gives;

s = (0² - 11.4²)/(2 × -3.8) = 17.1

The distance the required for the ranger's vehicle to com to rest, s = 17.1 (meters).

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B is incorrect because friction in air is too small to make an effect.

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