Answer:
V = 5.67 V
Q' = 2.09 x 10^-4 C
Explanation:
Potential difference, V = 5.67 V
Charge, Q = 3.49 x 10^-5 C
dielectric constant, K = 5.99
Capacitance of the capacitor
C = Q / V = 3.49 x 10^-5 / 5.67 = 6.15 x 10^-6 F
Now the dielectric is inserted, so,
C' = K x C = 5.99 x 6.15 x 10^-6 = 3.687 x 10^-5 F
After insertion of dielectric, as the battery is connected, so potential difference remains same.
Charge, Q' = C' x V = 3.687 x 10^-5 x 5.67 = 2.09 x 10^-4 C
