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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Question:

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.

Answers

Answer:The charge stored  is  [tex]Q = 4.25 *10^{-7 } \ C[/tex]The energy stored is  [tex]E = 2.55*10^{-6} \ J[/tex] Explanation:From the question we are told that     The area of the plates is  [tex]A = 0.40 \ m^2[/tex]      The separation between the plate is [tex]d = 0.10 \ mm =0.0001 \ m[/tex]       The potential difference is  [tex]V = 12 \ V[/tex]        The permitivity of free space is  [tex]\epsilon_o = 8.85 *10^{-12} C^2 \cdot N^{-1} \cdot m^2[/tex]          The dielectric constant of glass is K =  5.0Generally the capacitance of this capacitor is       [tex]C = \frac{\epsilon_o * A}{d}[/tex]substituting values        [tex]C = \frac{8.85*10^{-12} * 0.40}{0.0001}[/tex]         [tex]C = 3.34 *10^{-8} \ C[/tex]The charge stored is mathematically evaluated as          [tex]Q = CV[/tex]substituting values         [tex]Q = (3.54*10^{-8} * 12)[/tex]          [tex]Q = 4.25 *10^{-7 } \ C[/tex]The energy stored is            [tex]E = 0.5 * CV^2[/tex]substituting values          [tex]E = 0.5 * (4.25 *10^{-7} * 12^2)[/tex]            [tex]E = 2.55*10^{-6} \ J[/tex]

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