Question

11-15. Calculate the capacitance of two parallel plates of a capacitor with a separation VT distance of 10 mm and with an area of 2.5 X10• m2. (Please use Eo = 8.85 x 10-12F/m)

16-20. A parallel plate capacitor has a square plate with side measuring 6.0 cm and ..-•1 separated by a distance of 2 mm. Calculate the capacitance of this capacitor. (Please use E0 = 8.85 x 10-12F/m)

21-25. Calculate the potential difference of a battery connected to a capacitor consisting of two parallel plates with an area of 1.75 cm2 and a separation distance of 4.33 mm. Suppose that the charge present on the two plates is equal to 2.67 Pico coulombs. (Please use E0 = 8.85 x 10-12F/m)

26-30. A cylindrical capacitor has a length of 6 cm is made of two concentric rings with an V. inner radius of 2.5 cm and an outer radius of 3.5 cm. How much charge is present in this e capacitor if it is connected to a 15 V battery? (Please use Eo = 8.85 x 10-12F/m)

Answer 1

11-15. The capacitance is

$$C=\frac{\epsilon_0 A}{d}=2.21·10^{-7}\text{ F}.$$

(The substituted area of the plates is 2.5 X10² m²).

16-20. Using the same equation, we determine that

$$C=\frac{(8.85·10^{-12}) (0.06^2)}{0.002}=15.9\text{ pF}.$$

21-25. The potential difference is

$$V=\frac QC=\frac{Qd}{\epsilon_0A}=7.5\text{ V}.$$

26-30. The charge is

$$Q=VC.$$

The capacitance of the cylindrical capacitance is

$$C=\frac{2\pi\epsilon_0L}{\ln(R_2/R_1)},\\\space\\ Q=V\frac{2\pi\epsilon_0L}{\ln(R_2/R_1)}=1.49·10^{-10}\text{ C}.$$