The plates of a parallel plate capacitor each have an area of 0.40 m2 and are separated by a distance of 0.02 m. They are charged until the potential difference between the plates is 3000 V. The charged capacitor is then isolated.
How much work is required to move a –4.0 ìC charge from the negative plate to the positive plate of this system?
Work= force*distance= Eq*distance=Volt/distance*q*distance= Vq
So the work is Volts times charge.
thanks i though it was somnethin differnt cause they gave the distance and the area
W=deltaV*Q
W=(0-3000)*(-4*10^-6)
W=1.2*10^-2 J
No problem! It's always good to double-check and make sure you have the correct formula. In this case, you are correct that the work done to move a charge between the plates of a parallel plate capacitor is equal to the potential difference (Volts) multiplied by the charge (q). So, in this case, the work required to move a -4.0 μC charge from the negative plate to the positive plate would be:
Work = (potential difference) * (charge)
= (3000 V) * (-4.0 μC)
Make sure to convert the charge to Coulombs before calculating:
Work = (3000 V) * (-4.0 μC) / (1 μC / 10^(-6) C)
Calculating this, you would get:
Work = -12,000 μC * V / (1 μC / 10^(-6) C)
Simplifying,
Work = -12,000 V * 10^(-6) C
= -12 J
Therefore, the work required to move a -4.0 μC charge from the negative plate to the positive plate of this system is -12 Joules (J).
You've got it! The formula for calculating work in this case is W = V * q, where W is the work done, V is the potential difference (or voltage), and q is the charge being moved.
Given that the potential difference between the plates is 3000 V and the charge being moved is -4.0 μC, you can plug in these values into the formula:
Work (W) = V * q
= 3000 V * (-4.0 μC)
To complete the calculation, you'll need to convert the charge from microcoulombs (μC) to coulombs (C). Since 1 μC = 10^-6 C, multiply the charge by 10^-6:
Work (W) = 3000 V * (-4.0 μC * 10^-6)
= 3000 V * (-4.0 * 10^-6 C)
Finally, compute the value:
Work (W) = 3000 V * (-4.0 * 10^-6 C)
= -12 * 10^-3 J
= -0.012 J
The answer is -0.012 J, indicating that the negative work of -0.012 joules is required to move a -4.0 μC charge from the negative plate to the positive plate of the capacitor.