Answer:
The additional charge that will flow from the battery onto the positive plate is 0.924 nC
Explanation:
Given;
Area of the capacitor, A = 15.0 cm² = 15 x 10⁻⁴ m²
Separation distance, d = 3.00 mm = 3 x 10⁻³ m²
voltage of the capacitor, V = 58.0 V
dielectric constant, k = 4.60
Initial Capacitance of the capacitor before the addition of dielectric material:
[tex]C = \frac{\epsilon _oA}{d} = \frac{8.85*10^{-12}*15*10^{-4}}{3*10^{-3}} = 4.425 *10^{-12} F[/tex]
Initial charge across the parallel plates:
Q₁ = CV
Q = 4.425 x 10⁻¹² x 58 = 2.567 x 10⁻¹⁰ C
Capacitance of the capacitor after the addition of dielectric material:
Cequ. = Ck
Cequ. = 4.425 x 10⁻¹²F x 4.6 = 20.355 x 10⁻¹² F
Final charge across the parallel plates:
Q₂ = Cequ. x V
Q₂ = 20.355 x 10⁻¹² x 58 = 11.806 x 10⁻¹⁰ C
Additional charge = Q₂ - Q₁
= 11.806 x 10⁻¹⁰ C - 2.567 x 10⁻¹⁰ C
= 9.239 x 10⁻¹⁰ C
= 0.924 nC
Therefore, the additional charge that will flow from the battery onto the positive plate is 0.924 nC
