If you were to expend 10 J of work to push a 1 C charge against an electric field, what would be its change of voltage?
(b) When released, what will be the kinetic energy of the 1 C charge of the previous problem if it flies past its starting position?
To answer these questions, we need to understand the relationship between work, electric field, voltage, and kinetic energy.
(a) To calculate the change in voltage, we use the formula:
Work (W) = charge (Q) * voltage change (ΔV)
Given that the work (W) is 10 J and the charge (Q) is 1 C, we can rearrange the equation to solve for the voltage change (ΔV):
ΔV = W / Q
Substituting the values, we have:
ΔV = 10 J / 1 C
ΔV = 10 V
Therefore, the change in voltage would be 10 V.
(b) To calculate the kinetic energy, we can use the formula:
Kinetic energy (KE) = 1/2 * mass (m) * velocity^2
Given that the charge (Q) is 1 C, we can relate it to mass using the equation:
Q = charge (Q) = mass (m) * electric field (E)
Since the charge is 1 C, and the electric field is not given, we cannot directly calculate the mass. However, once the charge has been pushed against the electric field, it accumulates potential energy. When it is released, this potential energy is converted into kinetic energy, assuming no other forces are acting on it.
So, the change in potential energy will be equal to the kinetic energy. Hence, the kinetic energy of the 1 C charge will be equal to the work done on the charge, which is 10 J.
Therefore, the kinetic energy of the 1 C charge would also be 10 J.