Question

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 the potential difference is 200 v.

Answer 1

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 = $\frac{Q}{V}$

Further Explanation

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

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, 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

Answer 2

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.