Question

If I place a dielectric of thickness 2cm and relative permittivity 2 in between the plates of a parallel plate capacitor, with area $3c{m}^2$ and plate spacing 4 cm, what is the new capacitance?

• A. $2{\epsilon }_0\times {10}^{-2}F$
• B. ${\epsilon }_0\times {10}^{-2}F$
• C. $3{\epsilon }_0\times {10}^{-2}F$
• D. None of these.

Answer 1

The correct option is B ${\epsilon }_0\times {10}^{-2}F$

When a dielectric of thickness t is introduced, the capacitance changes to
$C=\frac{{\epsilon }_{0}A}{d-t+\frac{t}{k}}$
If you don’t know where this formula comes from, go back to the videos where this is derived.
Now, $C=\frac{{\epsilon }_0\times 3\times {10}^{-4}}{(4-2+\frac{2}{2})\times {10}^{-2}}$
$C={\epsilon }_0\times {10}^{-2}F$