A parallel plate capacitor with plate area A and plate separation $d=2 \mathrm{ m}$ has a capacitance of $4 \mu \mathrm{ F}$ . The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant $\mathrm{K}=3$ (as shown in figure) will be:

Question

A parallel plate capacitor with plate area A and plate separation

d=2 \mathrm{ m}

has a capacitance of

4 \mu \mathrm{ F}

. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant

\mathrm{K}=3

(as shown in figure) will be: A parallel plate capacitor with plate area A and plate separation d=2 m has a capacitance of 4 μ F . The new

Quantrex Academy · Accepted Answer

Correct answer: C. 6 μ F

We have, C_i=(A _0)/(d)=4 μ FC_f=(A _0)/(d-t+(t)/(k))=(A _0)/(d-(d)/(2)+(d)/(2 × 3))=(A ∈_0)/(d(1-(1)/(2)+(1)/(6)))=(4 μ F)/((2)/(3))=6 μ F

A parallel plate capacitor with plate area A and plate separation...