consider a parallel plate capacitor of capacity 10 microfarad with air filled in the gap between the plates, now one half of the space between the plates is filled with dielectric const 4 , the capacitor of plates changes to? please answer my question as soon as possible..

The capacitance of the parallel plate capacitor in terms of separation d, area A and permittivity is given as,
$C=\frac{{\epsilon }_{0}A}{d}$

when the parallel plate capacitor is half filled with dielectric and half filled with air then the change in the capacitance of the capacitor 'C' is given as,
 $C=C_1+C_2$
where C₁ is the capacitance of the capacitor half filled with air and C₂ is the capacitance of the capacitor filled with dielectric.

So, the net capacitance of the capacitor half filled with air and half filled with dielectric is given as,
 $C=\frac{{\epsilon }_0\left({\displaystyle \frac{1}{2}}A\right)}{d}+\frac{k{\epsilon }_0\left({\displaystyle \frac{1}{2}}A\right)}{d}$

as k for air is unity (1).

on solving the above question we get,
 $$\begin{gathered} C=\frac{{\epsilon }_{0}A}{2d}+\frac{k{\epsilon }_{0}A}{2d} \\ C=\frac{{\epsilon }_{0}A}{d}\left(\frac{1+k}{2}\right) \end{gathered}$$

It is given that the capacity of the capacitor is $10 \mu F$ and the dielectric constant is 4. So, on substituting the values in the equation we get,
$$\begin{gathered} C=\left(10\mu F\right)\left(\frac{1+4}{2}\right) \\ C=\left(10\mu F\right)\left(\frac{5}{2}\right) \\ C=\frac{50 \mu F}{2} \\ C=25 \mu F \end{gathered}$$