Question

Problems 4-9 refer to this same capacitor: The plates of a parallel plate capacitor each have an area of 0.40 m and are separated by a distance of 0.02 m. They are charged until the potential difference between the plates is 3000 V. The charged capacitor is then isolated so no charge is lost. Show how units work out. 4. Determine the magnitude of the electric field between the capacitor plates. A) 60 V/m B) 120 V/m C) 1.0 x 10s V/m D) 1.5 x 10 V/m E) 3.0 x 10 V/m 5. Determine the value of the capacitance. A) 9.0× 10-11 F B) 1.8× 10-10F C) 3.6-10-10F D) 4.8 x 10-10F E) 6.4× 10-10F 6. Determine the magnitude of the charge on either capacitor plate. A) 1.8 x 10 C B) 2.7 107 C C) 4.9 10"C D) 5.4 10 c E) 6.8 × 10-3 C 7. How much work is required to move a -4.0 uC charge from the negative plate to the positive plate of this system? A) -1.2 x 102J B) +1.2 x 10 J C)-2.4× 10-リ D) +2.4 x 102 E)-5.4× 10-2 J 8. Suppose that a dielectric sheet is inserted to completely fill the space between the plates and the potential difference between the plates drops to 1000 V. What is the capacitance of the system after the dielectric is inserted? A) 1.8 x 10t0 F B) 2.7 x 1010 F C) 5.4 x 10-10 F D) 6.2 x 10-10 F E) 6.8 10-10 F 9. Suppose that a dielectric sheet is inserted to completely fill the space between the plates and the potential difference between the plates drops to 1000 V. Determine the dielectric constant A) 0.333 B) 0.666 C) 3.0 D) 6.0 E) 2000

Answer 1

4. E = V / d = (3000 V) / 0.02 m

E = 1.5 x 10^5 V/m

Ans(D)

5. C = e0 A / d = (8.854 x 10^-12)(0.40)/0.02

C = 1.8 x 10^-10 F

Ans(B)

6. Q = C V = (1.8 x 10^-10)(3000) = 5.4 x 10^-7 C

Ans(D)

7. W = q delta(V) = (-4 x 10^-6)(0 - 3000)

W = 1.2 x 10^-3 J

Ans(B)

8. Q = C V = constant

C = (1.8 x 10^-10)(3000)/1000

C = 5.4 x 10^-10 F

Ans(C)

9. k = Cf/ Ci = 5.4/1.8 = 3

Ans(C)