A parallel plate capacitor filled with a medium of dielectric constant 10, is connected across a battery and is charged. The dielectric slab is replaced by another slab of dielectric constant 15. Then the energy of capacitor will :
A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be.
The charge on capacitor of capacitance 15$$\mu$$F in the figure given below is :
A parallel plate capacitor with plate area A and plate separation d = 2 m has a capacitance of 4 $$\mu$$F. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant K = 3 (as shown in figure) will be :