Answer:
Explanation:
a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV
C = capacitance of the capacitor (in Farads )
V = voltage (in volts) = 100V
C = ∈A/d
∈ = permittivity of free space = 8.85 × 10^-12 F/m
A = cross sectional area = 600 cm²
d= distance between the plates = 0.7cm
C = 8.85 × 10^-12 * 600/0.7
C = 7.59*10^-9Farads
Q = 7.59*10^-9 * 100
Q = 7.59*10^-7Coulombs
Q = 0.759*10^-6C
Q = 0.759µC
b) Energy stored in a capacitor is expressed as E = 1/2CV²
E = 1/2 * 7.59*10^-9 * 100²
E = 0.0000395Joules
E = 39.5*10^-6Joules
E = 39.5µJ
A) The charge Q on the positive plate of the capacitor is ; 0.759 µC
B) The energy stored in the capacitor increases by : 39.5 µJ
Given data :
Area of plates ( A ) = 600 cm²
Distance between plates ( d ) = 0.7 cm
Voltage across plates = 100 v
∈ ( permittivity of free space ) = 8.85 * 10⁻¹²
A) Determine the Charge on the positive plate of the capacitor
Q = CV --- ( 1 )
where ; C = ∈ * A / d and V = 100 v
∴ C = 8.85 * 10⁻¹² * 600 / 0.7 = 7.59 *10⁻⁹ F
Back to equation ( 1 )
Q = 7.59 *10⁻⁹ * 100
= 0.759 µC
B) Calculate how much The energy stored in the capacitor increases
E = 1/2 * C * V²
= 1/2 * 7.59 *10⁻⁹ * 100²
= 39.5 µJ
Hence we can conclude that The charge Q on the positive plate of the capacitor is ; 0.759 µC, The energy stored in the capacitor increases by : 39.5 µJ.
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