A 15-mu F air-filled capacitor is connected to a 50-V voltage source and becomes
ID: 1412462 • Letter: A
Question
A 15-mu F air-filled capacitor is connected to a 50-V voltage source and becomes fully charged. The voltage source is then removed and a slab of dielectric that completely fills the space between the plates is inserted. The dielectric has a dielectric constant of 5.0. What is the capacitance of the capacitor after the slab has been inserted? What is the potential difference across the plates of the capacitor after the slab has been inserted? What is the total potential energy stored within the capacitor?Explanation / Answer
Here ,
A) new capacitance , C = old capacitance * dielectric constant
new capacitance , C = 15 uF * 5
new capacitance , C = 75 uF
B)
as the charge on the capacitor will be contsant
the potential difference across the capacitor , V = initial potential/dielectric constant
potential difference across the capacitor , V = 50/5 = 10 V
C)
total potential energy stored = 0.5 * CV^2
total potential energy stored = 0.5 * 75 *10^-6 * 15^2
total potential energy stored = 8.44 *10^-3 J