X and Y are two capacitors(parallel plate) connected in series to a 12V supply.X has air between the plates and its capacitance is 10microfarads.Y contains a dielectric medium of K=5.calculate capacitance of Y and the potential difference between the plates of Y.

Assuming that the area and plate spacing is the same for the two capacitors, the capacitance of Y is 50 uF. In series, both will store the same charge Q but the Y capacitor will have 1/5 the voltage drop of X, since V = Q/C and C is five times greater for capacitor Y.

Vx + Vy = Vx + Vx/5 = 12
6/5 Vx = 12
Vx = 10 Volts
Vy = 2 Volts

oye thanku sooo much..nyc:)

To find the capacitance of Y, we can use the formula:

C = (K * C0)
Where C is the capacitance with dielectric, K is the dielectric constant, and C0 is the capacitance without dielectric.

In this case, the capacitance of Y with the dielectric medium (C) is what we need to find, the dielectric constant (K) is given as 5, and the capacitance without the dielectric (C0) is the capacitance of X.

Since X has air between the plates, its capacitance is given as 10 microfarads (10μF).

Therefore, we can calculate the capacitance of Y as:
C = (K * C0) = 5 * 10μF = 50μF

So, the capacitance of Y is 50 microfarads (50μF).

To find the potential difference between the plates of Y, we need to use the formula for capacitors connected in series:

1/C = 1/C1 + 1/C2

We know that C1 is the capacitance of X, which is 10μF, and C2 is the capacitance of Y, which is 50μF.

So, substituting the values into the formula:
1/C = 1/10μF + 1/50μF

To simplify the calculation, we can convert all the values to a common denominator:

1/C = (5+1)/50μF = 6/50μF

Now, taking the reciprocal of both sides:

C = 50μF/6

Thus, the potential difference between the plates of Y is 50μF/6 or approximately 8.33μF.