Two 11-cm-diameter electrodes 0.40 cm apart form a parallel-plate capacitor. The

Question

Two 11-cm-diameter electrodes 0.40 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 11 Vbattery.

A)What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes while the capacitor is attached to the battery? what is the Q, E, and deltaV.

B)What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 0.80 cm apart? The electrodes remain connected to the battery during this process. what is the Q E and deltaV

C) What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes after the original electrodes (not the modified electrodes of parts in B) are expanded until they are 22 cm in diameter while remaining connected to the battery? What is the Q, E, and deltaV

Explanation / Answer

multiple questions I will solve part 1) which again contains 3 sub parts.

1) Q = C * V = [o * A / d] * V = (8.854 x10^-12) * * (D / 2)² * V / d

Q = (8.854 x10^-12) * * (0.11 / 2)² * (11) / 0.0040

Q = 231.27 x10^(-12) coulombs -------------------->>>>>>>>>charge)

E = (V / d) = (15 / 0.0040) = 2750 V/m ------------>>>>>>>>>>>>electric field strength)

potential difference between the electrodes --------->>>>>>>>>>>>>>>> 11 v

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