Question

The capacitance of a parallel plate capacitor increases from 5 µf to 60 µf when a dielectric is filled between the plates. The dielectric constant of the dielectric is
(a) 65
(b) 55
(c) 12
(d) 10

Answer 1

Answer: (c) 12


The Capacitance of a Parallel Plate Capacitor is given by:

$\boxed{C=\frac{\varepsilon_{\circ}A}{d}}$

Where

$\varepsilon_{\circ}$ = Absolute Permittivity
$A$ = Area of Plates
$d$ = Distance between Plates


Here, We are given that the original capacitance was 5 $\mu F$
So, $C = 5 \, \, \mu F$


When a dielectric material is placed between the plates, the expression for capacitance becomes:

$\boxed{C = \frac{\varepsilon_{r}\varepsilon_{\circ}A}{d}}$

Here:
$\varepsilon_r$ = Relative Permittivity

It is also known as Dielectric Constant $K$

So, if we write new capacitance [when the dielectric is placed] as $C'$ then we have:

$C' = \frac{K\varepsilon_{\circ}A}{d} \\ \\ \\ \implies C' = KC$

We are also told that the new Capacitance is 60 $\mu F$

So we have:
$C' = KC \\ \\ \implies 60 = K \times 5 \\ \\ \implies K = \frac{60}{5} \\ \\ \implies \boxed{K=12}$
Thus, the dielectric constant is 12. So, the answer is Option (c).

Answer 2

the capacitance of a parallel plate capacitor increases from 5 uf to 60 uf. when a dielectric is filled between the plates. the dielectric constant of the dielectric is 12..........





your answer is =12



c