The electric field between the plates of a paper-separated (K = 3.75) capacitor is 9.21 104 V/m. The plates are 2.05 mm apart and the charge on each plate is 0.775 µC. Determine the capacitance of this capacitor.

Question

The electric field between the plates of a paper-separated (K = 3.75) capacitor is 9.21 104 V/m. The plates are 2.05 mm apart and the charge on each plate is 0.775 µC. Determine the capacitance of this capacitor.

Answer 1

E= 8.18•10⁴ V/m

d=2.2 mm
q=0.775 μ C
ε (paper)= 2.3
(a) C=q/U
E=U/d

C=q/Ed = ...
(b)
ε₀=8.85•10⁻¹² F/m

C= ε₀εA/d =>
A=Cd/ε₀ε=...

Answer 2

Well, isn't that "capaci-fun"! To determine the capacitance of this capacitor, we can use the formula C = Q / V, where C is the capacitance, Q is the charge on each plate, and V is the electric field between the plates.

Plugging in the values we have, the charge (Q) is 0.775 µC and the electric field (V) is 9.21 x 10^4 V/m.

So, the capacitance (C) is given by C = (0.775 µC) / (9.21 x 10^4 V/m).

But hold on a second! We have to convert the charge from microCoulombs to Coulombs first. Since there are 10^-6 C in 1 µC, the charge becomes 0.775 x 10^-6 C.

Substituting this value into our equation, we get C = (0.775 x 10^-6 C) / (9.21 x 10^4 V/m).

Simplifying a bit, we have C = 8.40 x 10^-12 F.

So, the clown-approved answer is that the capacitance of this capacitor is 8.40 picofarads (pF). Isn't that positively electrifying?

Answer 3

To determine the capacitance of the capacitor, you can use the formula:

C = Q / V

where C is the capacitance, Q is the charge on each plate, and V is the electric field.

In this case, the charge on each plate is given as 0.775 µC, which is equivalent to 0.775 × 10^(-6) C. The electric field is given as 9.21 × 10^4 V/m.

Plugging these values into the formula, we have:

C = (0.775 × 10^(-6) C) / (9.21 × 10^4 V/m)

Now, calculating the capacitance:

C = 0.775 × 10^(-6) C / 9.21 × 10^4 V/m

C = 0.084 µF

Therefore, the capacitance of this capacitor is 0.084 µF.

Answer 4

To determine the capacitance of the capacitor, we can use the formula:

C = Q / V

Where:
- C is the capacitance
- Q is the charge on each plate
- V is the voltage between the plates

Given:
Q = 0.775 µC (microCoulombs)
V = 9.21 × 10^4 V/m (Volts per meter)

We need to convert the charge from microCoulombs to Coulombs by dividing it by 10^6:

Q = 0.775 × 10^(-6) C

Now we can substitute these values into the formula to calculate the capacitance:

C = (0.775 × 10^(-6) C) / (9.21 × 10^4 V/m)

Simplifying:

C = 0.775 × 10^(-6) / 9.21 × 10^4

C = 8.41 × 10^(-11) F

Therefore, the capacitance of the capacitor is 8.41 × 10^(-11) Farads.