Question

C,Q,V and U are capacitance, charge, potential difference and energy stored in an isolated parallel 1 plate capacitor respectively. Now a dielectric slab is placed between plates of capacitor, the quantities that decreases

Answer 1

Answer:

C becomes kC and Q remains contant as there is no battery to increase or decrease charge ...and as CV = Q so c becomes k tyms so v becomes 1/k tyms as Q constant ...and U becomes 1/k tyms ....

so V and U decreased