Consider a parallel plate capacitor of 10 μ F (micro-farad) with air filled in the gap between the plates. Now one-half of the space between the plates is filled with a dielectric of dielectric constant 4, as shown in the figure. The capacity of the capacitor changes to

Question

Consider a parallel plate capacitor of 10 μ F (micro-farad) with air filled in the gap between the plates. Now one-half of the space between the plates is filled with a dielectric of dielectric constant 4, as shown in the figure. The capacity of the capacitor changes to (A) 25 μ F (B) 20 μ F (C) 40 μ F (D) 5 μ F

Tardigrade Admin · Accepted Answer

C1=(ε0((A/4))/d), C2=(K ε0((A/2))/d) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/b83184aaa0d9d0d4c73989420cf3b35a-.png" /> C3 =(ε0((A/4))/d) C text eq =C1+C2+C3 =((K+1/2)) (ε0 A/d) =((4+1/2)) × 10 =25 μ F

Q. Consider a parallel plate capacitor of $10 μ F$ (micro-farad) with air filled in the gap between the plates. Now one-half of the space between the plates is filled with a dielectric of dielectric constant $4$ , as shown in the figure. The capacity of the capacitor changes to

$25 μ F$

$20 μ F$

$40 μ F$

$5 μ F$

$C_{1} = \frac{ε _{0} ( \frac{A}{4} )}{d},$
$C_{2} = \frac{K ε _{0} ( \frac{A}{2} )}{d}$

$C_{3} = \frac{ε _{0} ( \frac{A}{4} )}{d}$
$C_{eq} = C_{1} + C_{2} + C_{3}$
$= (\frac{K + 1}{2}) \frac{ε _{0} A}{d}$
$= (\frac{4 + 1}{2}) \times 10$
$= 25 μ F$

Q. Consider a parallel plate capacitor of 10μF (micro-farad) with air filled in the gap between the plates. Now one-half of the space between the plates is filled with a dielectric of dielectric constant 4, as shown in the figure. The capacity of the capacitor changes to

25μF

20μF

40μF

5μF