A slab of copper of thickness b = 1.566 mm is thrust into a parallel-plate capacitor of C = 4.00×10-11 F of gap d = 10.0 mm, as shown in the figure; it is centered exactly halfway between the plates.
If a charge q = 3.00×10-6 C is maintained on the plates, what is the ratio of the stored energy before to that after the slab is inserted?
How much work is done on the slab as it is inserted?
Is the slab pulled in or must it be pushed in?
when you put the slab of copper in, you are making two capacitors in series, reducing capacitance.
Find the new capacitance. Then, knowing q is the same, find V across each capacitor.
Knowing Q, V you can find energy stored. Then, compare it to the original. If energy is greater, then you had to push the plate in, if energy is less, the copper was pulled in.
i think i know wat to do, but i'm having trouble finding the area.
How do you find the area?
To find the ratio of the stored energy before and after the slab is inserted in the parallel-plate capacitor, we need to compare the stored energy in the capacitor without the slab to the stored energy with the slab.
The formula for the energy stored in a parallel-plate capacitor is:
E = (1/2) * C * V^2
where E is the stored energy, C is the capacitance, and V is the voltage across the capacitor.
Before the slab is inserted, let's assume there is some voltage V1 across the capacitor. The energy stored in the capacitor without the slab is:
E1 = (1/2) * C * V1^2
After the slab is inserted, the capacitance of the capacitor changes due to the presence of the slab. Let's assume the new capacitance is C'. The voltage across the capacitor with the slab is V2. The energy stored in the capacitor with the slab is:
E2 = (1/2) * C' * V2^2
To find the ratio of the stored energy before to after, we can write:
(E1 / E2) = ( (1/2) * C * V1^2 ) / ( (1/2) * C' * V2^2 )
Since the slab is centered exactly halfway between the plates, it divides the gap d into two equal parts. So, the distance between the slab and each plate is d/2 = 10 mm / 2 = 5 mm.
To find the new capacitance C' with the slab inserted, we can use the formula:
(1/C') = (1/C) + (b / K * A)
where b is the thickness of the slab, K is the dielectric constant, and A is the area of the plates.
First, we need to find the area of the plates. Assuming the plates are rectangular, the area is:
A = l*w
where l is the length of the plates, which can be obtained from the given figure, and w is the width of the plates, which is the same as the gap d.
Now, let's calculate the values:
Area of the plates:
A = l * d = (length) * (gap) = (given figure)
New capacitance with the slab:
(1/C') = (1/C) + (b / K * A)
(C')^(-1) = (C)^(-1) + (b / K * A)
C' = 1 / ( (1/C) + (b / K * A) )
Once you find the new capacitance C', you can calculate the voltage V2 using the charge q and the new capacitance C':
V2 = q / C'
Finally, substitute the values of C, C', V1, and V2 into the (E1 / E2) formula to find the ratio of the stored energy before to after the slab is inserted.
To determine the work done on the slab as it is inserted into the capacitor, we can use the formula:
Work = q * V2
where q is the charge maintained on the plates and V2 is the voltage across the capacitor with the slab inserted.
Substitute the given values of q and V2 into the formula to find the work done on the slab.
Now, let's address whether the slab is pulled in or must be pushed in.
The slab will be pulled into the capacitor due to the electric force between the charges on the plates and the charges induced in the slab. As the capacitor gets charged, opposite charges get induced on the surfaces of the slab, creating an attractive force that pulls the slab into the capacitor. So, no external force is needed to insert the slab; it will be pulled in by the electric forces.