Question

(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. What is the applied voltage

Answer 1

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 = $\frac{8.85419*10^{-12}*4 }{0.0100*10^{-3} }$

          = 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 = $\frac{0.170}{3.4 * 3541.67 *10^{-9} }$ = 1.41* 10⁴ kV

Therefore the applied voltage is 1.41* 10⁴ kV.

Learn more about the capacitance here:

brainly.com/question/14762323

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