the elecric potential at a certain point is space is 12 V. What is the electric potential energy of a -3.0uc charge placed at that point?
is it +36uj?
PE= qV
Oh, I see you're "charged" up about this question! Well, to calculate the electric potential energy, we can use the formula:
Electric potential energy = charge × electric potential
In this case, the charge is -3.0 μC and the electric potential is 12 V. So let's do the math:
Electric potential energy = (-3.0 μC) × (12 V)
Multiplying these values, we get:
Electric potential energy = -36 μCV
So, the electric potential energy of the -3.0 μC charge at that point is indeed -36 μJ. However, I must clarify that the sign indicates the direction of the potential energy, not the magnitude, so it's not +36 μJ. Keep up the electrifying questions!
The electric potential energy of a charge in an electric field is given by the equation:
Electric Potential Energy = Charge × Electric Potential
In this case, the charge is -3.0 µC (microcoulombs) and the electric potential is 12 V (volts). To calculate the electric potential energy, we need to multiply the charge and the electric potential:
Electric Potential Energy = (-3.0 µC) × (12 V)
Now, let's perform the calculation:
Electric Potential Energy = -3.0 µC × 12 V
= -36 µC·V
= -36 µJ (microjoules)
So, the correct answer is -36 µJ, not +36 µJ.
To calculate the electric potential energy, we need to use the equation:
Electric Potential Energy (PE) = charge (q) * electric potential (V)
Given:
Charge (q) = -3.0 μC (microcoulombs)
Electric Potential (V) = 12 V
Now, let's substitute the given values into the equation:
PE = (-3.0 μC) * (12 V)
Ensure that the charge is in coulombs (C) instead of microcoulombs (μC):
1 μC = 10^-6 C
Therefore, -3.0 μC = -3.0 × 10^-6 C
PE = (-3.0 × 10^-6 C) * (12 V)
PE = -36 × 10^-6 C·V
PE = -36 μC·V
Hence, the electric potential energy is -36 μC·V, not +36 μC·V.