Answer :
Answer:
The capacitance is 90 pF.
Explanation:
Given that,
Distance between the plates= 1.50 mm
Charge on each plate = 0.0180μC
Potential difference = 200 V
Suppose we find the capacitance
We need to calculate the capacitance
Using formula of capacitance
[tex]Q=CV[/tex]
[tex]C=\dfrac{Q}{V}[/tex]
Where, Q = charge
C = capacitance
V = potential
Put the value into the formula
[tex]C=\dfrac{0.0180\times10^{-6}}{200}[/tex]
[tex]C=90\times10^{-12}[/tex]
[tex]C=90\ pF[/tex]
Hence, The capacitance is 90 pF.
The total energy used is 1.8 × 10⁻⁶ J
You can calculate the energy stored on a capacitor by using the following equation.
U = [tex]\frac{1}{2}[/tex]CV² , where C = [tex]\frac{Q}{V}[/tex]
Further Explanation
If you substitute C in the equation, then you can express the equation as
U = [tex]\frac{1}{2}[/tex]QV
The energy stored in the capacitor = U
The capacitance = C
The Voltage applied or potential differences is = V
From the question:
- The energy stored in the capacitor =?
- The distance between the two flat parallel plates = 1.50 mm
- The Voltage Applied = 200 V
- The magnitude of charge on each plate = 0.0180 μc = 1.80 × 10 ⁻⁸ C
To determine the potential energy of the capacitor, we use the formula
U = ½ QV
U = ½ x 1.80 × 10 ⁻⁸ x 200
U = 1.8 × 10⁻⁶ J
Therefore, the total energy stored 1.8 × 10⁻⁶ J
LEARN MORE:
- When the magnitude of the charge on each plate of an air-filled capacitor is 4 ?c, https://brainly.com/question/3912616
- A parallel-plate air capacitor with a capacitance of 260 pF has a charge of magnitude 0.155 μC on each plate https://brainly.com/question/12910809
KEYWORDS:
- air capacitor
- potential difference
- two parallel plates
- magnitude of charge
- voltage applied