please check these for me:
When a positive charge moves in the direction of the electric field, what happens to the electrical potential energy associated with the charge?
a. it increases
b. it decreases
c. it remains the same
d. it sharply increases, and then decreases
C
A parallel-plate capacitor has a capacitance of C F. If the area of the plates is doubled while the distance between the plates is halved, the new capacitance will be
a. 4C
b. 2C
c. C/2
d. C/4
A
The energy gained by electrons as they are accelerated by an electric field is
a. greater than the average loss in energy due to collisions
b. equal to the average loss in energy due to collissions
c. less than the average loss in energy due to collissions
d. not affected by the gain in energy due to collissions
D
Which set of information will allow you to calculate the kilowatt*hr usage?
a. the voltage and current in the circuit
b. the resistance, the current, and the time the circuit operates
c. the voltage and the resistence of the circuit
d. the current and the time the circuit operates
B
thanks for all the help!
#1: wrong
#2: correct
#3: wrong. Energy gained from E-field equals energy lost in collisions, at equilibrium
#4: correct
You're welcome! I'm glad I could help. Here's an explanation for each answer:
1. When a positive charge moves in the direction of the electric field, the electrical potential energy associated with the charge decreases. This is because the charge is moving in the same direction as the force exerted by the electric field, and work is done to move the charge. Since work is done against the electrical force, the electrical potential energy decreases. So, the correct answer is (b) it decreases.
2. The capacitance of a parallel-plate capacitor depends on the area of the plates and the distance between them. The formula for capacitance is C = ε₀(A/d), where C is the capacitance, ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates. In this case, if the area of the plates is doubled (2A) and the distance between the plates is halved (d/2), the new capacitance (C') can be calculated as C' = ε₀(2A/(d/2)). Simplifying this expression gives C' = 4C, where C is the initial capacitance. Therefore, the correct answer is (a) 4C.
3. The energy gained by electrons as they are accelerated by an electric field is greater than the average loss in energy due to collisions. This is because the electric field accelerates the electrons, increasing their kinetic energy. However, electron-electron collisions within the conductor result in some loss of energy. So, the correct answer is (a) greater than the average loss in energy due to collisions.
4. To calculate the kilowatt*hr usage, we need information about the power consumed by the circuit and the time the circuit operates. Power is calculated as P = IV, where P is power, I is current, and V is voltage. The kilowatt*hr usage can be obtained by multiplying the power (in kilowatts) by the time (in hours) the circuit operates. Therefore, the correct answer is (b) the resistance, the current, and the time the circuit operates.
I hope this helps! Let me know if you have any more questions.