Answer:
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
