Question

An engineer working in an electronics lab connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 240 V. Assume a plate separation of d 1.77 cm and a plate area of A-25.0 cm. when the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0
(a) Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before pC after Q pC
(b) Determine the capacitance (in F) and potential difference (in V) after immersion.
(c) Determine the change in energy (in nJ) of the capacitor nJ
(d) What If? Repeat parts (a) through (c) of the problem in the case that the capacitor is immersed in distilled water while still connected to the 240 V potential difference Calculate the charge on the plates (in pC) before and after the capacitor is submerged. (Enter the magnitudes.) before Q, - after Q Determine the capacitance (in F) and potential difference (in V) after immersion. pC pC Determine the change in energy (in nJ) of the capacitor. nJ

Answer 1

Answer:

a. 3.0 X 10^-10 C; 2.4 X 10 ^-8 C b. 1.0 X 10^-10 C; V= 240+3 =243 V c. 36 nJ  d. 12.2 nJ

Explanation:

a. C=Q/V = ε0εrA/d => Q=εoεrAV/d; since dielectric constant for free space is 8.85 x 10^-12 and that of distilled water is 80, this shows that other dielectric material is free space with εr=1, Q=5.21 x 10^-15 C; εr for water is 80;

c. Energy, E =1/2(QV)