Question

A parallel plate capacitor of capacitance,$C$ (without dielectrics) is filled by dielectric slabs as shown in figure.Then the new capacitance of the capacitor is

• A. $39C$
• B. $4C$
• C. $34C$
• D. $3C$

Answer 1

The correct option is B $4C$

Ref. Image I

Ref. Image II $C_1=\frac{K{\epsilon }_{0}A}{\frac{d}{2}}$

$=2\times 6\times \frac{{\epsilon }_{0}A}{d}$

$=12C$

$C_2=\frac{K_2{\epsilon }_0\frac{A}{2}}{\frac{d}{2}}=2C$

$C_3=\frac{K_3{\epsilon }_0\frac{A}{2}}{\frac{d}{2}}=4\frac{{\epsilon }_{0}A}{d}=4C$

$C_{eq}=C_2+C_3$ Ref. image III

$=4C+2C$

$=6C$

$C_{eq}=\frac{12C\times 6C}{12C+6C}=\frac{72C^2}{18C}$

$C_{eq}=4C$