A parallel plate capacitor filled with mica having εr = 5 is connected to a 10 V battery. The area of the parallel plate is 6 m^2 and separation distance is 6 mm. Find the capacitance.*1 point4.43 × 10^-8 F4.43 × 10^-9 F44.3 × 10^-8 F443 × 10^-9 F
Question
A parallel plate capacitor filled with mica having εr = 5 is connected to a 10 V battery. The area of the parallel plate is 6 m^2 and separation distance is 6 mm. Find the capacitance.*1 point4.43 × 10^-8 F4.43 × 10^-9 F44.3 × 10^-8 F443 × 10^-9 F
Solution
The capacitance (C) of a parallel plate capacitor can be calculated using the formula:
C = ε0 * εr * A / d
where: ε0 is the permittivity of free space (8.85 × 10^-12 F/m), εr is the relative permittivity (also known as the dielectric constant), A is the area of one of the plates (in square meters), and d is the separation distance between the plates (in meters).
Given: εr = 5, A = 6 m^2, and d = 6 mm = 6 × 10^-3 m.
Substituting these values into the formula gives:
C = (8.85 × 10^-12 F/m) * 5 * (6 m^2) / (6 × 10^-3 m) = 4.43 × 10^-9 F.
So, the capacitance of the capacitor is 4.43 × 10^-9 F.
Similar Questions
Now the friends try a homework problem.A capacitor is constructed with two parallel metal plates each with an area of 0.74 m2 and separated by d = 0.80 cm. The two plates are connected to a 4.0-volt battery. The current continues until a charge of magnitude Q accumulates on each of the oppositely charged plates.Find the electric field in the region between the two plates. V/mFind the charge Q. CFind the capacitance of the parallel plates.
A parallel-plate capacitor of surface area of 10cm2 and plate separation distance (d) of 3mm is connected to a 15V battery as shown in the following figure. Find the work done to decrease the plate separation distance to 1mm while the capacitor is connected to the battery.
A parallel plate capacitor has circular plate of 0.1m radius and 0.2cm separation. Calculate the capacitance1 point0.000000000139 F0.000000000159 F0.00000000139 F000000000139 F
A parallel-plate capacitor is formed of of two 20.0 cm × 20.0 cm plates spaced 2.50 cm apart. The plates are charged to ±±4.00 nC. What is the magnitude of the electric field between the plates?
Solve the given problem below. Submit your illustration and solutions as PDF format as an attachment to your final answer (60 points). A parallel plate capacitor is constructed of two (2) square conducting plates with side length l = 10.0 cm. The distance between the plates is d = 0.250 cm. A dielectric with a dielectric constant k = 15.0 and thickness 0.250 cm is inserted between the plates. The dielectric is 10.0 cm wide and is 5.0 cm long.Provide an illustration that best represents the scenario (20 points).Derive and provide the formula that shall best address the given scenario (20 points).Determine the capacitance of this system (20 points).
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.