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RoseWind [281]
3 years ago
8

An air capacitor is made from two flat parallel plates 1.50 mm apart. The magnitude of charge on each plate is 0.0180 μc when th

e potential difference is 200 v.
Physics
2 answers:
Masja [62]3 years ago
7 0

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 = \frac{1}{2}CV² , where C = \frac{Q}{V}

<h2>Further Explanation</h2>

If you substitute C in the equation, then you can express the equation as

U = \frac{1}{2}QV

The energy stored in the capacitor = U

The capacitance = C

The Voltage applied or potential differences is = V

<u>From the question: </u>

  • 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, 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 brainly.com/question/12910809

KEYWORDS:

  • air capacitor
  • potential difference
  • two parallel plates
  • magnitude of charge
  • voltage applied
Cloud [144]3 years ago
5 0

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

Q=CV

C=\dfrac{Q}{V}

Where, Q = charge

C = capacitance

V = potential

Put the value into the formula

C=\dfrac{0.0180\times10^{-6}}{200}

C=90\times10^{-12}

C=90\ pF

Hence, The capacitance is 90 pF.

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