Design a 4.50e-7 C Capacitor w/ 220V & 0.375 m2 Plates

[purduegirl]

Homework Statement

You must design a capacitor capable of storing 4.50 × 10-7 C of charge, but you only have a 220- V power supply and two metal plates of area 0.375 m2 each. What is the largest value the plate separation can be so that you can accomplish the task?

Homework Equations

V = Q*d/A*epilson 0

The Attempt at a Solution

I solved for d, and got 1.62E-2 m. Again, don't know what I am doing wrong here.

 

[alphysicist]

Hi purduegirl,

That's not quite the answer I get from your equation. Why don't you calculate it again; if you still get the same thing, please show what equation you got when you solved for d and then exactly the numbers you plugged in your calculator.

 

[Moetasim]

I would approach this problem by first understanding the basic principles of capacitors. A capacitor is a device that stores electrical energy in an electric field between two conductive plates. The amount of charge that a capacitor can hold is directly proportional to the voltage applied and the surface area of the plates, and inversely proportional to the distance between the plates.

In this case, we are given a charge of 4.50 × 10-7 C, a voltage of 220 V, and a plate area of 0.375 m2. Using the equation V = Q*d/A*epilson 0, we can solve for the plate separation (d). However, before we do that, we need to convert our units to SI units.

First, we convert the charge to coulombs (C) by multiplying it by 10^-7. This gives us a charge of 4.50 × 10^-14 C. Next, we convert the plate area to square meters (m2) by multiplying it by 10^-4. This gives us a plate area of 3.75 × 10^-5 m2.

Now, we can plug these values into our equation: 220 = 4.50 × 10^-14 * d / (3.75 × 10^-5 * 8.85 × 10^-12). Solving for d, we get a plate separation of 1.62 × 10^-2 m, which is the same value you obtained.

Therefore, the largest value the plate separation can be is 1.62 × 10^-2 m in order to store 4.50 × 10^-7 C of charge with a voltage of 220 V and plate area of 0.375 m2. This solution is in line with the basic principles of capacitors and shows that the equation we used is correct.