A parallel-plate capacitor has plates with an area of 250 cm2 separated by 1.0 mm. What is the electric field between the plates when they are charged with 7.000×10-7 C?
Solution:
Using the formula,
=> E = Q/(e0 * A)
=> E = 7 * 10^-7 /(8.85 * 10^-12 * 250 * 10^-4)
=> E = 7 * 10^9 /(8.85 * 250)
=> E = 3.16 * 10^6 N/C
Please comment for queries please rate
Thanks
A parallel-plate capacitor has plates with an area of 250 cm2 separated by 1.0 mm. What...
An air-filled parallel-plate capacitor has plates of area 2.50 cm2 separated by 3.00 mm. The capacitor is connected to a 21.0-V battery. (a) Find the value of its capacitance. (b) What is the charge on the capacitor? (c) What is the magnitude of the uniform electric field between the plates?
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.60 mm. The capacitor is connected to a(n) 17.0 V battery. (a) Find the value of its capacitance. _____pF (b) What is the charge on the capacitor? _____pC (c) What is the magnitude of the uniform electric field between the plates? ______N/C
An air-filled parallel-plate capacitor has plates of area 2.70 cm2 separated by 0.50 mm. The capacitor is connected to a 6.0-V battery. (a) Find the value of its capacitance. ________ pF (b) What is the charge on the capacitor? ________ pC (c) What is the magnitude of the uniform electric field between the plates? ________ V/m
An air-filled parallel-plate capacitor has plates of area 2.10 cm2 separated by 1.00 mm. The capacitor is connected to a 24.0-V battery. (a) Find the value of its capacitance. pF (b) What is the charge on the capacitor? pC (c) What is the magnitude of the uniform electric field between the plates? V/m
A)A parallel-plate capacitor has an area of 5.40 cm2 and the plates are separated by 0.800 mm. The capacitor stores a charge of 386.0 pC. What is the potential difference across the plates of the capacitor? The permittivity of a vacuum is 8.8542 × 10−12 C 2/N · m2 Answer in units of V. B) What is the magnitude of the uniform electric field in the region that is located between the plates? Answer in units of V/m.
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 has plates with area 0.0225 m2 separated by 1.00 mm of Teflon. (a) Calculate the charge on the plates when they are charged to a potential difference of 12.0 V. (b) Use Gauss’s law (Eq. 24.23) to calculate the electric field inside the Teflon. (c) Use Gauss’s law to calculate the electric field if the voltage source is disconnected and the Teflon is removed
A 200 pF parallel-plate capacitor has ±35 nC charges on its plates. The plates are separated by 0.3 mm. Find: a) the area of each plate; cm2 b) the potential difference between the plates; V c) the electric field between the plates; N/C
1. Two parallel plates, each of area 5.17 cm2, are separated by 4.90 mm. The space between the plates is filled with air. A voltage of 4.75 V is applied between the plates. Calculate the magnitude of the electric field between the plates. 2. Calculate the amount of the electric charge stored on each plate. 3. Now distilled water is placed between the plates and the capacitor is charged up again to the same voltage as before. Calculate the magnitude...