Question

A dielctric of dielectric constant 2.5 is filled in the gap between the plates of a capacitor. Calculate the factor by which the capacitance is increased, if the dielectric is only sufficient to fill up one fourth of the gap

Answer 1

Capacitor capacity until full

$C=\frac{\epsilon_0 \cdot S}{d}$

Capacitor capacity after filling

$C^{'}=\frac{C_1 \cdot C_2 }{C_1 + C_2}$

Where

$C_1=\frac{\epsilon\cdot \epsilon_0 \cdot S}{\frac{1}{4} \cdot d}=\epsilon\cdot \frac{4\cdot \epsilon_0 \cdot S}{ d}=4 \cdot \epsilon \cdot C$

$C_2=\frac{ \epsilon_0 \cdot S}{\frac{3}{4} \cdot d}=\frac{4\cdot \epsilon_0 \cdot S}{3 \cdot d}=\frac{4}{3} \cdot C$

Then write

$C^{'}=\frac{C_1 \cdot C_2 }{C_1 + C_2}=\frac{4 \cdot \epsilon \cdot C \cdot \frac{4}{3} \cdot C}{4 \cdot \epsilon \cdot C + \frac{4}{3} \cdot C}=\frac{4 \cdot \epsilon \cdot \frac{4}{3} \cdot C}{4 \cdot \epsilon + \frac{4}{3} }=\frac{16 \cdot \epsilon }{12 \cdot \epsilon + 4 } \cdot C=k \cdot C$

Then the desired coefficient is

$k=\frac{16 \cdot \epsilon }{12 \cdot \epsilon + 4 }=\frac{16 \cdot 2.5 }{12 \cdot 2.5 + 4 }=\frac{40 }{34}=1.176$

Capacitor capacity will increase 1.176 times