# Math Help - capacitance 2

1. ## capacitance 2

two plates of a parallel plate capacitor are 0.01m apart a dielectric slab of dielectric constant 6 and thickness 0.005 m is introduced between the plates parallel to the plates of the capacitor determine the distance between the plates such that the capacitance remains same after suitable adjusments of the plates

2. The capacitance in the first part is given by:

$C = \frac{\epsilon_o \epsilon_r A}{d} = \frac{(\epsilon_o)(\epsilon_r)(A)}{0.01}$

The capacitance in the second part is given by:

$C = \frac{(\epsilon_o)(6)(A)}{0.005} + \frac{(\epsilon_o)(\epsilon_r) A}{d}$

Since they are equal;

$\frac{(\epsilon_o)(\epsilon_r)(A)}{0.01} = \frac{(\epsilon_o)(6)(A)}{0.005} + \frac{(\epsilon_o)(\epsilon_r) A}{d}$

$\frac{\epsilon_r}{0.01} = \frac{6}{0.005} + \frac{\epsilon_r}{d}$

I'm guessing that what was in between the plates prior to the insertion od the dielectric is vacuum, so epsilon nought becomes 1.

You solve for d, the thickness of the vacuum left, then 0.005+d is your final answer.