(a) A certain parallel-plate capacitor has plates of area 4.00m2 , separated by 0.0100 mm of nylon, and stores 0.170 C of charge
1 answer:
The applied voltage is 1.41* 10⁴ kV.
Calculation of applied voltage-
The capacitance of parallel plates in the absence of a dielectric is:
C = ∈₀A/d
where, "d" = the spacing between the plates = 0.01 * 10⁻³ m
"A" = the area of the plates = 4 m²
"∈₀" = the constant = 8.85419*10⁻¹² C²/Nm²
So, C₀ =
= 3541.67 *10⁻⁹ F
The dielectric constant, κ= 3.4, is defined as:
κ = C/C₀
The effective capacitance (with the dielectric) is denoted by "C", whereas the original capacitance is denoted by "C₀". (without the dielectric).
As a result, the new capacitance is:
C = κC₀
However, capacitance is connected to a voltage by:
C = Q/V
utilizing the increased capacitance and accounting for "V":
κC₀ = Q/V
V = Q/κC₀ = = 1.41* 10⁴ kV
Therefore the applied voltage is 1.41* 10⁴ kV.
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