Question

Print Your Full Name A dielectric slab is slowly inserted the capacitor is connected to a battery. As it is being inserted: s. between the plates of a parallel plate capacitor while (A) The capacitance, the potential difference between the plates, and the charge on the positive plate all increase. (B) The capacitance, the potential difference between the plates, and the charge on the (C) The potential difference between the plates increases, the charge on the positive plate (D) The capacitance and the charge on the positive plate decrease, but the potential (E) The capacitance and the charge on the plate increase, but the potential difference positive plate all decrease. decreases, and the capacitance remains the same difference between the plates remains the same. between the plates remains the same. 9. Each ofthe four capacitors shown is 500μ, and the voltmeter reads 1000V The magnitude of the charge on each capacitor plate is: (A) 0.2 C (B) 0.5 C (C) 20 C (D) 50 C (E) None of the above 10. A portion of a circuit is shown, with the values of the currents given for some branches. What is the direction and value of the current i? 4A 12A 13A 5 A T2A (A) ↓, 6A (B) ↑, 6A (C) ↓,4A (D) ↑,4A (E) ↓, 2A

Answer 1

8) Option D is the correct answer.

When capacitor is being connected to the battery. Potential diffrence across the capacitor reamins constant.
And that value is equal to potential diffrence across the terminals of the battery.

But capacitance of the capacitor increases. so, charge on the capacitor also increases.

Before insering dielectric.

potential difference across capacitor = V

Capacitnace = C

charge = Q

After insering dielectric.

potential difference across capacitor = V

Capacitnace = k*C

charge = k*Q

9) B) 0.5 C

Q = C*V

= 500*10^-6*1000

= 0.5 C

10) A) downward, 6A (using Kirchoff's current law)