A charge of 0.25 C moved from a position where the electric potential is 20 V to a position where the potential is 50 V. What is the change in potential energy?
work= PE1-PE= q1V1-q2V2=.25 (-30)=-7.5J
PE change = - work done
Why did the charge move from the position where the electric potential is 20 V to the position where it's 50 V? Did it hear there was a "Power Surge Party" happening there? As for the change in potential energy, it can be calculated using the formula ∆U = q∆V, where ∆U is the change in potential energy, q is the charge, and ∆V is the change in potential. So in this case, the change in potential energy would be 0.25 C * (50 V - 20 V) = 7.5 J. Voila!
To find the change in potential energy, we need to use the formula:
∆PE = q * ∆V
where:
∆PE is the change in potential energy,
q is the charge, and
∆V is the change in electric potential.
In this case, the charge (q) is 0.25 C, and the change in electric potential (∆V) is the difference between the final potential (50 V) and the initial potential (20 V).
∆V = 50 V - 20 V
∆V = 30 V
Substituting these values into the formula, we get:
∆PE = 0.25 C * 30 V
∆PE = 7.5 J
Therefore, the change in potential energy is 7.5 Joules.
To find the change in potential energy, we can use the formula:
Change in Potential Energy = q * ΔV
where q is the charge and ΔV is the change in electric potential.
Given:
q = 0.25 C
ΔV = (final potential) - (initial potential) = 50 V - 20 V = 30 V
Substituting these values into the formula:
Change in Potential Energy = 0.25 C * 30 V
Now, we can calculate the change in potential energy:
Change in Potential Energy = 0.25 C * 30 V
Change in Potential Energy = 7.5 Joules
Therefore, the change in potential energy is 7.5 Joules.