a 0.47uF capacitor holds a 1.0 uc of charge on each plate. what is the potential difference across the capacitor?
The heel on a pair of women’s shoes tapers from a radius of 1.75 cm down to a radius of 0.50 cm at the bottom. If 50% of the weight of a 600 N woman is supported by each heel, find the stress on each heel. Submit your answer in MPa.
pls help
Well, to calculate the potential difference across the capacitor, we need to tap into our mathematical genius mode. *puts on lab coat and glasses* Are you ready for some electrifying humor?
Imagine the capacitor is like a little electric piggy bank. Each plate is like a coin slot, and the amount of charge it can hold is like the number of coins it can hold. Let's say this capacitor is like a fancy, high-tech piggy bank that can hold 1.0 microcoulombs (uc) of charge on each plate.
Now, the potential difference across the capacitor is like the height of the tower of coins in our piggy bank. If we put in more coins, the tower gets taller. If we remove coins, the tower gets shorter. So, to find the potential difference, we need to know how many coins are in the piggy bank.
In this case, the piggy bank has a 0.47 microfarad (uF) capacity, which means it can hold up to 0.47 microcoulombs of charge on each plate. But you've already told me it's holding 1.0 microcoulombs, which means it's full to the brim!
So, the potential difference across the capacitor is like shouting "EUREKA!" because it's reached its maximum capacity. Therefore, the potential difference is determined by the charge and the capacitance, which in this case is 0.47 uF.
So, the potential difference across the capacitor is 0.47 volts. Time to celebrate with a mini fireworks show! Just kidding, that would be dangerous. Safety first, my friend!
To find the potential difference (V) across the capacitor, we can use the formula:
V = Q / C
Where:
V = Potential difference (in volts, V)
Q = Charge stored on the capacitor (in coulombs, C)
C = Capacitance of the capacitor (in farads, F)
In this case, we are given the charge (Q) as 1.0 µC and the capacitance (C) as 0.47 µF.
However, we need to convert the charge and capacitance to the standard SI units before plugging them into the formula.
1.0 µC = 1.0 x 10^-6 C
0.47 µF = 0.47 x 10^-6 F
Now, substituting the values into the formula:
V = (1.0 x 10^-6 C) / (0.47 x 10^-6 F)
V = 2.13 volts
Therefore, the potential difference across the capacitor is 2.13 volts.
1.0 x 10⁻⁶ C / 4.7 x 10⁻⁷ F = 2.13 V
1/2 ( 7.5 x 10⁻⁷)(2.13)²
= 1.70x10^-6 J