Question

Two parallel plates, each having area A = 3490cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.36cm. You may assume (contrary to the drawing) that the separation between the plates is small compared with either linear dimension of the plate. A dielectric having dielectric constant κ = 3.6 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 3490 cm2 and thickness equal to half of the separation (= 0.18 cm) . What is the charge on the top plate of this capacitor?

Answer 1

To find the charge on the top plate of this capacitor, use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. First, find the capacitance using the formula C = κε₀A/d. Then, substitute the capacitance and the voltage into the formula Q = CV to find the charge.

Step-by-step explanation:

To find the charge on the top plate of this capacitor, we can use the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. First, let's find the capacitance.

The capacitance of a parallel plate capacitor with a dielectric can be calculated using the formula C = κε₀A/d, where C is the capacitance, κ is the dielectric constant, ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.

Substituting the given values into the formula, we have C = 3.6 * 8.85 * 10^-12 * 3490 / 0.36.

Now, we can substitute the capacitance and the voltage into the formula Q = CV to find the charge. The charge on the top plate of the capacitor is 6 * (3.6 * 8.85 * 10^-12 * 3490 / 0.36).

Answer 2

So the problem ask to find the charge that is present at the top plate of the capacitor where as the two plates has an area of 3490cm^2. Base on that data and the connection of the circuits or plates, i came up with an answer of 8*10^-7 coulomb. I hope you are satisfied with my answer and feel free to ask for more