Question

A laptop battery has an emf of 10.8 v. the laptop uses 0.70 a while running. (a) how much charge moves through the battery each second? (b) by how much does the electric potential energy of this charge increase as it moves through the battery?

Answer 1

The charge that moves through the battery each second is 0.70 coulombs. The electric potential energy of this charge increases by 7.56 joules as it moves through the battery.

Step-by-step explanation:

The question relates to electric charge and electric potential energy in the context of a laptop battery with an electromotive force (emf) of 10.8 volts that uses 0.70 amperes (A) of current while running. To find out how much charge moves through the battery each second, we use the relationship Q = It, where Q is the charge in coulombs, I is the current in amperes, and t is the time in seconds.

Since we are looking for the charge per second, we have Q = (0.70 A)(1 s) = 0.70 C. As for the increase in electric potential energy, we use the formula ΔPE = QV, where ΔPE is the change in electric potential energy, Q is the charge, and V is the potential difference or emf of the battery. This yields ΔPE = (0.70 C)(10.8 V) = 7.56 joules.

Answer 2

The laptop battery moves 0.70 C of charge each second, and the electric potential energy increases by 7.56 J per second as this charge moves through the battery with an emf of 10.8 V.

Step-by-step explanation:

Understanding Laptop Battery Charge and Electric Potential Energy

To answer the first part of your question: (a) how much charge moves through the battery each second? We use the relationship between current and charge, which is I = Q/t, where I is the current in amperes, Q is the charge in coulombs, and t is the time in seconds. Given that the laptop uses a current of 0.70 A, we calculate the charge per second (which in this case, time t is 1 second) as Q = I * t = 0.70 C/s.

For the second part: (b) by how much does the electric potential energy of this charge increase as it moves through the battery? The electric potential energy change (ΔU) is given by ΔU = Q * emf, where emf is the electromotive force. With an emf of 10.8 V and a charge Q of 0.70 C passing through each second, the increase in electric potential energy is ΔU = 0.70 C * 10.8 V = 7.56 J per second.