A parallel-plate capacitor with plates of area 600 cm^2 is charged to a potential difference V and is then disconnected from the voltage source. When the plates are moved 0.7 cm farther apart, the voltage between the plates increases by 100 V.(a) What is the charge Q on the positive plate of the capacitor?nC(b) How much does the energy stored in the capacitor increase due to the movement of the plates?µJ

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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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