A capacitor is made up of two metal plates, each having an area of 0.02 m2 and spaced 2.5 mm apart. A potential difference of 250 V across the plates establishes a charge of 50 nC. Calculate:

a) The electric field strength,
b) The electric flux density,
c) The absolute permittivity of the dielectric,
d) The relative permittivity if the dielectric.

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Hi gfhgd

To calculate the values you mentioned, we can use the following formulas and steps:

a) The electric field strength (E) is given by the formula:

E = V / d

where V is the potential difference and d is the distance between the plates.

Plugging in the values: V = 250 V, d = 2.5 mm = 0.0025 m, we can calculate E.

E = 250 / 0.0025
= 100,000 N/C

Therefore, the electric field strength is 100,000 N/C.

b) The electric flux density (D) is calculated using the formula:

D = Q / A

where Q is the charge and A is the area of one of the plates.

Plugging in the values: Q = 50 nC = 50 x 10^-9 C, A = 0.02 m^2, we can calculate D.

D = (50 x 10^-9) / 0.02
= 2.5 x 10^-6 C/m^2

Therefore, the electric flux density is 2.5 x 10^-6 C/m^2.

c) The absolute permittivity of the dielectric (ε0) can be calculated using the formula:

ε0 = (Q / (A * εr)) / E

where Q is the charge, A is the area of one of the plates, E is the electric field strength and εr is the relative permittivity.

Since the charge and area remain the same, we can solve for ε0.

ε0 = (50 x 10^-9) / (0.02 * εr * 100,000)

Therefore, to find ε0, we need to know the value of εr.

d) The relative permittivity (εr) can be found using the formula:

εr = ε0 / εr

Rearranging the formula, we have:

εr = (50 x 10^-9) / ((0.02 * 100,000) / ε0)

To solve for εr, we need to know the value of ε0, which we can find using the formula in part c.

Therefore, we cannot calculate the value of the relative permittivity (εr) without knowing the value of ε0.

Please provide the value of ε0 in order to calculate the value of εr.