Capacitor Exercise (part 3)

My solution, just for fun.

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import math

Num = 10**6

K1 = 4.0
K2 = 3.0
d = 10.0
e0 = 1.0
a = 100.0

e1 = K1*e0
e2 = K2*e0

dx = a/Num
dA = a*dx

############################

C = 0.0

x = dx

while x <= a - dx:

    y = math.sqrt(x)

    dC1 = e1*dA/(d-y)
    dC2 = e2*dA/y

    dC = dC1*dC2/(dC1 + dC2)

    C = C + dC

    x = x + dx

############################

print Num
print C

#>>> 
#100000
#3286.82579219
#>>> ================================ RESTART ================================
#>>> 
#1000000
#3286.88428375
#>>> 

Consider a small element from a distance $x$ from the left side of the given diagram of thickness $dx$ . This can be treated as two elementary parallel plate capacitors in series.

$dC_1 = \frac{a \ dx \ K_2 \epsilon_o}{y}$ $dC_2 = \frac{a \ dx \ K_1 \epsilon_o}{d-y}$

Since these two elementary capacitors are in series, its equivalent capacitance is:

$dC = \frac{dC_1 \ dC2}{dC_1 + dC_2}$ $\implies dC = \frac{a \epsilon_o K_1 K_2 \ dx}{K_2(d-y) + K_1 y}$

Now, $y=\sqrt{x}$ , and so:

$dC = \frac{a \epsilon_o K_1 K_2 \ dx}{K_2(d-\sqrt{x}) + K_1 \sqrt{x}}$

Plugging in numbers and simplifying gives:

$dC = \frac{1200 \ dx}{30 + \sqrt{x}}$

$C = 1200 \int_{0}^{100} \frac{dx}{30 + \sqrt{x}} \approx 3286.89$