A parallelplate capacitor has capacitance 260 F If the dista

A parallel-plate capacitor has capacitance 26.0 F. If the distance between the plates is tripled, the plate area is tripled and a dielectric with dielectric constant 7.00 is placed between the plates, what will the capacitance of the new capacitor be?

Solution

Capacitance of a capacitor is given by

C=e_oA/d

Given

d_new=3d

and K=7 (dieletric inserted)

=>C_new/C =(Ke_oA/d_new)/(e_oA/d) =(K/3d)/(1/d)

C_new = (K/3)C =(7/3)*26

C_new=60.67 uF