first we need to calculate the capacitance of the parallel plate capacitor,
C = (ε * A)/d
where C is the capacitance, A is the area of plate and d is the separation between plates.
C = 8.85 * 10^-12 * 1.5/(1.4 * 10^-3)
C = 9.48 * 10^-9 Farad
Now, expression for the charge stored in a capacitor is,
Q = C*V
Q = 9.48 * 10^-9 * 2.7 * 10^3 C
Q = 25.60 μC
The charge stored in the capacitor is 25.60 μC.
A parallel plate capacitor has metal plates, each of area 1.50 m2, separated by 1.40 mm....
A parallel plate capacitor has metal plates, each of area 1.40 m2, separated by 2.00 mm. What charge (in µC) is stored in this capacitor if a voltage of 2.20 ✕ 103 V is applied to it?
A parallel plate capacitor has metal plates, each of area 1.25 m2, separated by 1.20 mm. What charge (in µC) is stored in this capacitor if a voltage of 2.50 ✕ 103 V is applied to it?
A parallel plate capacitor has metal plates, each of area 1.25 m2, separated by 1.10 mm. What charge (in µC) is stored in this capacitor if a voltage of 2.30 ✕ 103 V is applied to it?
a parallel plate capacitor had metal plates, each of area 1.5 A parallel applied to it? plate capacitor has metal plates, each of area 1.50 m2, t charge (in pC) is stored in this capacitor if a voltage of 2.70 × 102 Vis
The plates of a parallel-plate capacitor each have an area of 0.1 m2 and are separated by a 1 mm thick layer of glass. The capacitor is connected to a 11 V battery. (The dielectric constant for glass is 5.) a) Find the capacitance. b) Find the charge stored. c) Find the electric field between the plates.
1.Find the capacitance of a parallel plate capacitor having plates of area 5.00 m2 that are separated by 0.200 mm of Teflon. 2.What capacitance is needed to store 3.00 µC of charge at a voltage of 120 V?
A parallel plate capacitor is constructed with plate area of 0.80 m2 and a plate separation of 0.10 mm. When it is charged to a potential difference of 12 V, the charge stored on it is = micro C. A parallel plate capacitor is constructed with plate area of 0.40 m2 and a plate separation of 0.10 mm. When it is charged to a potential difference of 12 V, the charge stored on it is= micro C. A parallel-plate capacitor...
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
A parallel plate capacitor is constructed with plate area of 0.40 m2 and a plate separation of 0.10 mm. How much charge is stored on it when it is charged to a potential difference of 12 V? A parallel-plate capacitor is filled with air, and the plates are separated by 0.050 mm. If the capacitance is 17.3 pF, what is the plate area? A parallel plate capacitor is constructed with plate area of 0.0010 m2 and a plate separation of...
Each plate of a parallel-plate air-filled capacitor has an area of 0.007 m2, and the separation of the plates is 0.06 mm. An electric field of 3.95 × 106 V/m is present between the plates. What is the surface charge density on the plates? (ε0 = 8.85 × 10-12 C2/N ∙ m2) (Give your answer to the nearest 0.1 µC/m2).