A 15-mu F air-filled capacitor is connected to a 50-V voltage source and becomes

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

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