A43 v battery charges a 4750 pf air gap capacitor. after disconnecting. porcelain (k-7) is placed between the plates. what is the resulting charge of the plates? o 154 hc o 144 ac o 1.73 hc o 1.23 ac s

1.2255 × 10⁻⁶ μC

Explanation:

Given

Capacitance, C = 4750 pF = 4750×10⁻¹² F

Potential difference, V = 43 V

Relative permittivity of the dielectric, k = 7

now we know

charge Q = CV

⇒Q = 4750×10⁻¹² × 43 = 2.0425 × 10⁻⁷ C

Now, the charge (Q') due to the placing of the dielectric in between the gap

Q' = C'V

where,

C' = kC = 7 ×  4750×10⁻¹² = 33250 × 10⁻¹² F

substituting the C' in Q'

we get

Q' = 33250 × 10⁻¹² × 43 = 1.42975 × 10⁻⁶ C

Thus,

Resulting charge will be = Q' - Q

⇒Resulting charge = 1.42975 × 10⁻⁶ C - 2.0425 × 10⁻⁷ C = 1.2255 × 10⁻⁶ C

Resulting charge =1.2255 × 10⁻⁶ μC

amplitude

hope this