A parallel-plate capacitor has plate area of 0.12 m2and plate separation 1.2 cm. It is charged by a battery to potential difference of 120 V, then disconnected. A dielectric slab, thickness 4.0 mm and dielectric constant 4.8, is placed symmetrically between the plates.(a) Calculate the capacitance before and after the slab is inserted.(b) Calculate the free charge q before and after the slab is inserted.(c) Calculate the magnitude of the electric field in the space between the plates and the dielectric, and in the dielectric itself.(d) With the slab in place, calculate the potential difference across the plates.(e) How much external work is required to insert the slab (Please explain all the steps)
A parallel-plate capacitor has plate area of 0.12 m2and plate separation 1.2 cm. It is charged...
A parallel-plate capacitor has plates of area 0.13 m2 and a separation of 1.9 cm. A battery charges the plates to a potential difference of 150 V and is then disconnected. A dielectric slab of thickness 9.3 mm and dielectric constant 4.0 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...
A parallel-plate capacitor has a plate area of A = 250 cm2 and a separation of d = 2.00 mm. The capacitor is charged to a potential difference of V0 = 150 V by a battery. A dielectric sheet (κ = 3.50) of the same area but thickness ℓ = 1.00 mm is placed between the plates without disconnecting the battery. (See figure 24-18 on page 642). Determine the initial capacitance of the air-filled capacitor. Determine the charge on the...
A parallel-plate capacitor has a plate area of A = 250 cm2 and a separation of d = 2.00 mm. The capacitor is charged to a potential difference of V0 = 150 V by a battery. A dielectric sheet (κ = 3.50) of the same area but thickness ℓ = 1.00 mm is placed between the plates without disconnecting the battery. (See figure 24-18 on page 642). Determine the electric field in the dielectric. Determine the free charge on the...
A parallel-plate capacitor of capacitance Co, plate area A, spacing d is charged to voltage V. and then disconnected from the charging battery. A slab with dielectric constant K and thickness d/2 is thrust into the capacitor, as shown in the figure below; the slab is exactly halfway between the plates. к (a) What is the new capacitance in terms of Co? (b) What is the ratio of the stored energy before to that after the slab is inserted (U/0.)?...
The plates of an air-filled parallel-plate capacitor with a plate area of 16.5 cm2 and a separation of 8.80 mm are charged to a 130-V potential difference. After the plates are disconnected from the source, a porcelain dielectric with κ = 6.5 is inserted between the plates of the capacitor. (a) What is the charge on the capacitor before and after the dielectric is inserted? Qi = ___C Qf = ____C (b) What is the capacitance of the capacitor after...
The figure shows a parallel-plate capacitor of plate area A and plate separation d. A potential differenceV0 is applied between the plates. While the
battery remains connected, a dielectric slab of thickness b and dielectric constant κ is placed between the plates
as shown. Assume A = 130 cm2, d = 1.94
cm, V0 = 72.6 V, b = 0.735 cm, and κ =
3.15. Calculate (a) the capacitance,(b) the charge on the capacitor plates,(c) the electric field in the gap, and(d)...
A parallel plate capacitor of capacitance Co has plates of area A with separation d between them. When it is connected to a battery of voltage Vo, it has a charge of magnitude Qo on its plates. It is then disconnected from the battery and the space between the plates is filled with a material of dielectric constant 3. After the dielectric is added, the magnitudes of the charge on the plates and the potential difference between them are 15.
A parallel-plate capacitor in air has a plate separation of 1.35 cm and a plate area of 25.0 cm². The plates are charged to a potential difference of 270 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. (a) Determine the charge on the plates before and after immersion before pC after pC (b) Determine the capacitance and potential difference after immersion F V AV, (c) Determine the change...
A parallel plate capacitor of capacitance Co has plates of area A with separation d between them. When it is connected to a battery of voltage Vo, it has charge of magnitude Qo on its plates. It is then disconnected from the battery and the plates are pulled apart to a separation 2d without discharging them. After the plates are 2d apart, by what factor does the capacitance change? By what factor does the potential difference between the plates change?
Which of the following would increase the capacitance of a parallel-plate capacitor? I. Insert a dielectric between the plates. II. Increase the surface area of each plate. III. Increase the separation distance between the plates. O I and II only OII and III only All of the above. A capacitor is charged with a battery to a voltage V and then disconnected from the battery. A dielectric is inserted between the plates. When the dielectric is inserted, what happens to...