Question

Two 13.0-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 12 V battery. Part A) What is the charge on each electrode while the capacitor is attached to the battery? Part B) What is the electric field strength inside the capacitor while the capacitor is attached to the battery? Part C)What is the potential difference between the electrodes while the capacitor is attached to the battery? Part D) What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.1 cm apart? The electrodes remain connected to the battery during this process. Part E)What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each other until they are 1.1 cm apart? The electrodes remain connected to the battery during this process. Part F)What is the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.1 cm apart? The electrodes remain connected to the battery during this process.

Answer 1

capacitance of the capacitor C = E_0A/d = E_0 pi r^2/d..r = 6.5em...d = 0.5em = 2.35 times 10^-11 F = 23.5PF change in each electrode Q = ev = 2.35 times 10^-11 times 12 = 2.82 times 10^-10 C (B) assuming that electric field is uniform between the electrodes E = dv/dx = 12/5.0 times 10^-3 = 2.4 times 10^3 N/C (+ plate to - plate) (C) the potential difference is 12v (d) when the electrodes are 1.1 cm apart c^dash = E_0A/d^dash = E_0 pi r^2/d^dash = 1.068 times 10^-11 F Q^dash = ev = 1.28 times 10^-10C e) electric field E^dash = dv/dx = 12/1.1 times 10^-2 = 1.09 times 10^3 N/C f) potential difference will be 12v