Physics, 01.07.2020 15:01 mi364

An air-filled parallel-plate capacitor is constructed with a plate area of 0.40 m2 and a plate separation of 0.10 mm. It is then charged to a potential difference of 12 V? (ε0 = 8.85 × 10-12 C2 /N ∙ m2 ) (a) How much charge is stored on each of its plates? (b) If the capacitor is then filled with a glass dielectric (K = 5.0), by how much does the charge if it changes at all? The capacitor is not connected to a battery.

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Answer from: khikhi1705

The charge stored is $Q = 4.25 *10^{-7 } \ C$

The energy stored is $E = 2.55*10^{-6} \ J$

Explanation:

From the question we are told that

The area of the plates is $A = 0.40 \ m^2$

The separation between the plate is $d = 0.10 \ mm =0.0001 \ m$

The potential difference is $V = 12 \ V$

The permitivity of free space is $\epsilon_o = 8.85 *10^{-12} C^2 \cdot N^{-1} \cdot m^2$

The dielectric constant of glass is K = 5.0

Generally the capacitance of this capacitor is

$C = \frac{\epsilon_o * A}{d}$

substituting values

$C = \frac{8.85*10^{-12} * 0.40}{0.0001}$

$C = 3.34 *10^{-8} \ C$

The charge stored is mathematically evaluated as

Q = CV

substituting values

$Q = (3.54*10^{-8} * 12)$

$Q = 4.25 *10^{-7 } \ C$

The energy stored is

E = 0.5 * CV^2

substituting values

$E = 0.5 * (4.25 *10^{-7} * 12^2)$

$E = 2.55*10^{-6} \ J$

Answer from: Quest

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Answer from: Quest

Sorry i don’t know what to do

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