A parallel plate capacitor of capacitance 200 µF is connected to a battery of 200 V. A dielectric slab of dielectric constant 2 is now inserted into the space between plates of capacitor while the battery remain connected. The change in the electrostatic energy in the capacitor will be J.
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- A parallel-plate capacitor has plates of area 0.11 m2 and a separation of 1.1 cm. A battery charges the plates to a potential difference of 130 V and is then disconnected. A dielectric slab of thickness 3.7 mm and dielectric constant 9.0 is then placed symmetrically between the plates. (a) What is the capacitance before the slab is inserted? (b) What is the capacitance with the slab in place? What is the free charge q (c) before and (d) after the slab is inserted? What is the magnitude of the electric field (e) in the space between the plates and dielectric and (f) in the dielectric itself? (g) With the slab in place, what is the potential difference across the plates? (h) How much external work is involved in inserting the slab? (a) Number i Units (b) Number i Units (c) Number i Units (d) Number Units (e) Number Units (f) Number UnitsThe capacitor is charged to a voltage of 4.00 kv using a power source that is then removed. The gap between the plates is then filled by a dielectric layer. The charge on each plate stays constant at 2.50 kv despite the reduction in the potential difference between the plates. Calculate the initial capacitance value of the system.A parallel plate capacitor with plate separation do is charged by connecting it to a battery. The energy stored by the capacitor is Ug. The battery is then disconnected and the plate separation is changed to d1 = 0.7do. If U1 is the energy stored by the capacitor when the plate separation is di, what is the ratio U1/Uo? O (a) 0.7 O (b) 0.837 O (c) 0.49 O (d) 2.04 O (e) 1.43 O (f) 1.2
- An air filled capacitor C= 10uF is connected to a 15V battery. The capacitor is disconnected from the battery and the distance between the plates is doubled. What is the energy stored in the capacitor now?A parallel plate capacitor with a capacitance of 0.86 F with a has plate area A and are separated by a distance d. It is charged up to 11 V. The capacitor is now disconnected from the battery. If the area of the plates increases by a factor of 3 and the separation decreases by a factor of 10, find the new charge on the capacitor in Coulombs.In a parallel-plate capacitor, the two plates each have an area of 0.460 m2 and are spaced 3.00 mm apart.The capacitor is charged to a voltage of 4.00 kV using a power source that is then removed. The gap between the plates is then filled by a dielectric layer. The charge on each plate stays constant at 2.50 kV despite the reduction in the potential difference between the plates.Calculate the initial capacitance value of the system.