Answer to Question #95756 in Electric Circuits for Faris Halim

Question

A capacitor with double the area and double the dielectric thickness will have.

Option A. double the capacitance.
Option B. the same capacitance.
Option C. half the capacitance.

Answer 1

The original capacitance is given by

$C_{o}=k\frac{\varepsilon_{o}A_{o}}{d_{o}}$

Where

• Dielectric constant $K$
• Parallel plate area $A_{o}$
• distance between more plates $d_{o}$

Considering two dielectrics of constanta K (equal) and width of $do$

Each half can be considered a capacitor, if it is assumed that the plates have an electric charge + Q and -Q, the electric field in each capacitor is:

$E_{1}=\frac{Q}{k\varepsilon_{o} *2A_{o}} \\ E_{2}=\frac{Q}{k\varepsilon_{o} *2A_{o}}$ remember that you have twice the area Ao

The potential difference between the plates is equal to:

remember that each capacitor has a thickness do

$V=E_{1}d_{o}+E_{2}d_{o} \\ V=\frac{Q}{k\varepsilon_{o} *2A_{o}}+\frac{Q}{k\varepsilon_{o} *2A_{o}} \\ V=2\frac{Q}{k\varepsilon_{o} *2A_{o}} \\V=\frac{Q}{k\varepsilon_{o} *A_{o}}$

The new capacitance is equal to

$C=\frac{Q}{V} \\ C=\frac{Q}{\frac{Q}{k\varepsilon_{o} *A_{o}}}$

Simplifying the new capacitance is$C=k\varepsilon_{o}A_{o}$

Comparing with the original capacitance

$\frac{C}{C_{o}}=\frac{k\varepsilon_{o}A_{o}}{k\frac{\varepsilon_{o}A_{o}}{d_{o}}}\\\frac{C}{C_{o}}=1 \\C=C_{o}$

Solution:$\boxed{\text{Option B. the same capacitance.}}$