A parallel plate capacitor is given a definite potential difference. Keeping the potential difference same, a slab of thickness 3 mm is placed between the plates. To do this, the distance between the plates is increased by 2.4 mm. Calculate the dielectric constant of the slab.

10
15
5
8

Question

A parallel plate capacitor is given a definite potential difference. Keeping the potential difference same, a slab of thickness 3 mm is placed between the plates. To do this, the distance between the plates is increased by 2.4 mm. Calculate the dielectric constant of the slab.

10
15
5
8

Toppr Admin · Accepted Answer

-x3B5-Ad-x3B5-0Ad-x2032-x2212-t-r-x3B5-rord-d-x2032-x2212-t-t-x3B5-rr-x2212-r-x3B5-r-d-x2032-x2212-d-2-4mmr-1-x2212-1-x3B5-r-2-41-x2212-1-x3B5-r-2-43-x21D2-x3B5-r-306-5

A parallel plate capacitor is given a definite potential difference. Keeping the potential difference same, a slab of thickness 3 mm is placed between the plates. To do this, the distance between the plates is increased by 2.4 mm. Calculate the dielectric constant of the slab.101558

εAd=ε0Ad′−t+rεrord=d′−t+tεrr−rεr=d′−d=2.4mmr(1−1εr)=2.41−1εr=2.43⇒εr=306=5

A parallel plate capacitor is given a definite potential difference. Keeping the potential difference same, a slab of thickness 3 mm is placed between the plates. To do this, the distance between the plates is increased by 2.4 mm. Calculate the dielectric constant of the slab.

10
15
5
8

$\frac{ε A}{d} = \frac{ε_{0} A}{d^{'} - t + \frac{r}{ε_{r}}}$
or
$d = d^{'} - t + \frac{t}{ε_{r}}$
$r - \frac{r}{ε_{r}} = d^{'} - d = 2.4 m m$
$r (1 - \frac{1}{ε_{r}}) = 2.4$
$1 - \frac{1}{ε_{r}} = \frac{2.4}{3} \Rightarrow ε_{r} = \frac{30}{6} = 5$