A parallel plate capacitor is partially filled with a dielectric constant k=5. then
Answers
The complete question is:
A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be:
Given:
Capacitance of the capacitor = C
Dielectric constant when it is half filled = 5
To find:
Percentage increase in the capacitance.
Solution:
Initial capacitance= ∈0A/ d
where d is the distance between the places,
A is the area
When it is half filled by a dielectric of dielectric constant k then
C1 = k∈0A/ (d/2) = 2k∈0A/ d
and C2 = ∈0A/ (d/2) = 2∈0A/ d
Resultant capacitance C (capacitors are divided in series) :
1/C = 1/C1 + 1/C2
= 1/2∈0A/ d+ 1/2k∈0A/ d
= d/2∈0A + d/2k∈0A
Putting the value of k = 5
1/C = (d/2∈0A)*(6/5)
C = 5∈0A/ 3d
Hence, % increase in capacitance is = ((5∈0A/3d- ∈0A/d ) / ∈0A/d) *100
= 2/3* 100 = 66.67%
Therefore the percentage increase in the capacitance is 66.67%.
