Want this question answered?
119 joules per coulombCharges don't get joules as they flow through a circuit. They lose them.Every coulomb of charge that flows through a circuit ... from one terminal of a119-volt power supply, around the circuit, and back to the other terminal ...loses 119 joules during the trip.
That depends on the voltage. In general, a coulomb of charge will either gain or lose (depending on the direction) one joule of energy for every volt of potential difference. For example, if the battery has 12 V, a coulomb of charge will gain or lose 12 joules of energy when going from one terminal to the other.
A positive test charge of 1.6 x 10-11 C is placed in an electric field The force acting on it is 3.2 x 10-4 N What is the magnitude of the electric field intensity at the point where the charge is placed
By definition a volt is a joule per coulomb, V=W/Q (V is voltage, W is work done or energy measured in joules, Q is charge measured in coulombs) therefore 1 volt is 1 joule per 1 coulomb of charge (1C of charge is a very large amount to expect to see very small numbers for charge)
Coulombs for charge or Joules for heat capacity
119 joules per coulombCharges don't get joules as they flow through a circuit. They lose them.Every coulomb of charge that flows through a circuit ... from one terminal of a119-volt power supply, around the circuit, and back to the other terminal ...loses 119 joules during the trip.
it would be 10 joules because all you do is divide 10 joules by 1 coulomb of charge and you get 10 joules or (V) volts
That depends on the voltage. In general, a coulomb of charge will either gain or lose (depending on the direction) one joule of energy for every volt of potential difference. For example, if the battery has 12 V, a coulomb of charge will gain or lose 12 joules of energy when going from one terminal to the other.
A joule/coulomb is represented by the volt. Example: a 9v battery provides 9 joules of energy to every coulomb of charge that passes through it.
A positive test charge of 1.6 x 10-11 C is placed in an electric field The force acting on it is 3.2 x 10-4 N What is the magnitude of the electric field intensity at the point where the charge is placed
By definition a volt is a joule per coulomb, V=W/Q (V is voltage, W is work done or energy measured in joules, Q is charge measured in coulombs) therefore 1 volt is 1 joule per 1 coulomb of charge (1C of charge is a very large amount to expect to see very small numbers for charge)
You need to multiply the number of coulombs by the number of volts. If the two batteries are in series, then you can add the voltage of both batteries.
The potential difference ('voltage') is equal to the work done per unit charge, i.e. the energy given to each Coulomb of charge. So, a six Volt battery provides six Joules of energy to each Coulomb of charge.
A Coulomb is the unit of charge. It is a fundamental unit, representing the number of elementary charges (typically, electrons) available to do work. Its numerical value is about 6.241510x1018 elementary charges Important combined units based on the coulomb are the ampere, which is coulombs per second, the volt, which is joules per coulomb, and the volt-ampere, which is joules per second, or watts.
Voltage is "electrical pressure", so to speak, or energy per charge. Volts is joules per coulomb.
Coulombs for charge or Joules for heat capacity
The source of electrons in a circuit is supplied by an electric potential difference across two points . This potential difference in a circuit is called as voltage and is measured in joules per coulomb or volts.