A 5.0 µC point charge is moved within an electric field and has an electric potential energy change of 10.0 J. What is the electric potential difference before and after the charge was moved? Show work. (µC = 1.0 × 10–6
Vq=energy
so delta V must be 10 volts.
What do I do NOW?
You mean V * q = energy
so
V = energy / q
The potential difference * charge = potential energy change of charge
delta V = 10J/(5*10^-6) = 2*10^6 volts
Well, now that we know the change in electric potential energy is 10.0 J and the charge is 5.0 µC, we can use the formula V = ΔU/q to calculate the electric potential difference. So, we have V = 10.0 J / 5.0 × 10–6 C.
Now, let's do some math.
First, let's convert the charge from microcoulombs to coulombs. 5.0 µC is equal to 5.0 × 10–6 C.
Now we can substitute the values into the formula: V = 10.0 J / (5.0 × 10–6 C).
By simplifying the equation, we get V = 2.0 × 10^6 J/C.
So, the electric potential difference before and after the charge was moved is 2.0 × 10^6 volts.
Hope that helps, and remember, volts are like the electric version of dad jokes - they always have a lot of energy!
To find the electric potential difference, also known as the voltage (ΔV), you can use the equation:
ΔV = ΔPE / q
where ΔPE is the change in electric potential energy and q is the charge.
In this case, the charge q is 5.0 µC, which is equal to 5.0 x 10^-6 C. The change in electric potential energy ΔPE is given as 10.0 J.
Substituting these values into the equation, we get:
ΔV = 10.0 J / (5.0 x 10^-6 C)
Simplifying the expression, we have:
ΔV = (10.0 J) / (5.0 x 10^-6 C)
Now, let's perform the arithmetic to calculate the electric potential difference:
ΔV = 2.0 x 10^6 V
Therefore, the electric potential difference before and after the charge was moved is 2.0 x 10^6 volts.
To find the electric potential difference before and after the charge was moved, you need to divide the electric potential energy change by the charge of the point charge.
In this case, the electric potential energy change is 10.0 J and the charge of the point charge is 5.0 µC, which can be written as 5.0 × 10^-6 C.
So, the electric potential difference (ΔV) can be calculated using the formula ΔV = ΔU / q, where ΔU is the change in electric potential energy and q is the charge.
ΔV = ΔU / q
ΔV = 10.0 J / (5.0 × 10^-6 C)
Now, divide 10.0 J by 5.0 × 10^-6 C:
ΔV = 10.0 J / (5.0 × 10^-6 C) = 2.0 × 10^6 V
Therefore, the electric potential difference before and after the charge was moved is 2.0 × 10^6 V (or 2.0 MV).