Question

You are given a parallel plate capacitor that has plates of area 29 cm2 which are separated by 0.0100 mm of nylon (dielectric constant κ = 3.4). You are told to assume that this capacitor stores 0.18 mC of charge.Calculate what the applied voltage in kV should be for this capacitor.

Answer 1

Answer:

20.6kV

Explanation:

The capacitance (C₀) of a parallel plate capacitor is given by;

C₀ = A x ε₀ / d

Where;

A = Area of one of the plates = 29cm² = 0.0029m²

ε₀ = permittivity of free space = 8.85 x 10⁻¹²F/m

d = distance between the plates = 0.0100mm = 0.00001m

But since there is a dielectric material (nylon) between the plates, the effective capacitance (C) will increase to k x C₀.  i.e

C = k x C₀   -----------------------(i)

Where;

k = dielectric constant = 3.4 and;

C₀ = A x ε₀ / d

Substitute the value of C₀ into equation (i) to give;

C = k x A x ε₀ / d  -----------------(ii)

Substitute the values of k, A, d and C₀ into equation (ii) to give;

C = 3.4 x 0.0029 x 8.85 x 10⁻¹² / 0.00001

C = 8.7261 x 10⁻⁹ F

From the capacitance (C) calculated above, the amount of voltage(V) supplied can be found using the following relation;

Q = C x V         ------------------------------(iii)

Where Q is the charge on the capacitor = 0.18mC = 0.18 x 10⁻³C

Make V the subject of the formula in equation (iii) above;

V = Q / C

Substitute the values of Q and C into the equation to give;

V = (0.18 x 10⁻³) / (8.7261 x 10⁻⁹)

V = 0.0206 x 10⁶V

V = 20600V

V = 20.6kV

Therefore, the voltage applied in kV is 20.6

Answer 2

Answer:

20.60 kV

Explanation:

Capacitance of parallel plates without dielectric between them is:

$C=\frac{\varepsilon_{0}A}{d}$

with d the distance between the plates, A the area of the plates and ε₀ the constant $8.85419\times10^{-12}\frac{C^{2}}{Nm^{2}}$, so :

$C_0=\frac{(8.85419\times10^{-12})(0.0029)}{0.0100\times10^{-3}}=2.57\times10^{-9} F$

But the dielectric constant is defined as:

$k=\frac{C}{C_{0}}$

With C the effective capacitance (with the dielectric) and Co the original capacitance (without the dielectric). So, the new capacitance is:

$C=kC_0$

But capacitance is related with voltage by:

$C=\frac{Q}{V}$

with Q the charge and V the voltage, using the new capacitance and solving for V:

$kC_0=\frac{Q}{V}$

$V=\frac{Q}{kC_0}=\frac{0.18\times10{-3}}{(3.4)(2.57\times10^{-9})}=20599.68V=20.60 kV$