How much work does the electric field do in moving a proton from a point with a potential of +120 V to a point where it is -60 V? Express your answer in joules.
I've looked through my notes, and I'm sure the answer is really easy, but I'm not seeing it.
I know that Work(Wba)= PEa - PEb and that
Electrical Potential Vba = (Vb - Va)/q where q is the charge.
I'm sorry, I think i've just confused myself and I keep trying to relate equations but it's not working.
No problem, I'm here to help you figure it out step-by-step!
To calculate the work done by the electric field in moving a proton, we can use the equation:
Work (W) = ΔPE (change in electric potential energy)
The change in electric potential energy can be calculated using the formula:
ΔPE = q * ΔV
Where:
- ΔPE is the change in electric potential energy
- q is the charge of the particle (in this case, the charge of a proton is +1.6 x 10^-19 C)
- ΔV is the change in potential (Vb - Va)
Given:
- ΔV = -60 V - (+120 V) = -180 V
- q = +1.6 x 10^-19 C
Now we can substitute these values into the equation to find the change in electric potential energy:
ΔPE = (1.6 x 10^-19 C) * (-180 V)
Calculating this expression gives us:
ΔPE = -2.88 x 10^-17 J
The negative sign indicates that work was done by the electric field. Therefore, the work done is 2.88 x 10^-17 joules.
No problem! Let's break it down step by step.
To calculate the work done by the electric field, we need to use the formula:
Work (W) = Change in Potential Energy (ΔPE).
In this case, the change in potential energy can be calculated using the equation:
ΔPE = q * ΔV
where q is the charge of the proton and ΔV is the change in electrical potential.
Given that the potential at point A is +120 V and the potential at point B is -60 V, the change in potential (ΔV) can be calculated as:
ΔV = VB - VA = -60 V - 120 V
ΔV = -180 V
Now, we need to determine the charge of the proton. The charge of a proton is typically expressed as +e, where e is the elementary charge. The elementary charge is approximately 1.6 x 10^-19 Coulombs.
So, the charge (q) of the proton can be written as:
q = +e = +1.6 x 10^-19 C
Now we can substitute the values into the formula for ΔPE:
ΔPE = q * ΔV = (1.6 x 10^-19 C) * (-180 V)
Calculate the product of the charge and the change in potential to find the ΔPE:
ΔPE = -2.88 x 10^-17 J
The work done by the electric field (W) equals the change in potential energy (ΔPE), so the answer is:
W = -2.88 x 10^-17 J
Note that the negative sign indicates that the work is done on the proton since it moves from a higher potential to a lower potential.
Your notes are wrong on the second equation
Work=Force*distance= Eqd= (Va-Vb) q
The difference in potential is 180volts, q is the charge on an electron.