If 6.40 x 10^-19 joule of work is required to move a proton between two points A and B in an electric field, what is the potential difference between points A and B?
Given:
W = 6.40 x 10^-19
The answer is 4.00V
Energy difference (J) = V * Q
= (Potential difference, in Volts) x (Charge, in Coulombs)
Solve for the potentila difference
Well, I guess the potential difference between points A and B is so shocking that it requires 6.40 x 10^-19 joules of work to move a proton! That's electrifying!
To find the potential difference between points A and B, we can use the formula:
W = q * V
where W is the work done on the charge, q is the charge of the particle, and V is the potential difference.
In this case, we are given the work done (W) as 6.40 x 10^-19 joules. Since the charged particle is a proton, we know that the charge (q) is the elementary charge, which is +1.6 x 10^-19 coulombs.
Substituting these values into the formula, we have:
6.40 x 10^-19 = (+1.6 x 10^-19) * V
To isolate V, we divide both sides of the equation by +1.6 x 10^-19:
V = (6.40 x 10^-19) / (1.6 x 10^-19)
Simplifying the expression, we get:
V = 4
Therefore, the potential difference between points A and B is 4 volts.