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Disc Capacitors
Load capacitor question ...?

2 disc diameter of 2.1 centimeters 1.5 millimeters to form spaced intervals in a parallel plate capacitor. The electric field between the disks is 4.4 × 10 ^ 5 W / m. How What is the charge on each disk? Electrons from the negative plate Starts. At a rate of 2.5 × 10 in ^ 7m / s strikes the positive plate What is E-negative rates were at plate do? BC 660 W Numbers and found the voltage across the capacitor then abandoned.

Voltage is correct. The fee for the disc, most Straight forward to the information given to the capacitance of the capacitor Calculation and the definition of capacitance: the C = Q / Veronica Performed using two geometric plates separated by a very small distance, The limits are described in, for you by the capacitance (As given in the infinite plate capacitor can approximate the geometry Vacuum, in the absence of dielectric, the formula of Gauss's law of induction Use), the C = epsilon_0 * / d and epsilon_0 is the permittivity of free space can be done. And d of the area between the two plates are separating. Then, the capacitance To calculate the capacitance and voltage, you know, each plate fee Can be found (ie, question and response). As for the second part of the question Energy conservation issues. The charging voltage and the potential energy : U = q * and V. So, if you remember to say that e ,, Then the potential energy of the E * V on the positive plate Negative plate potential energy is 0. Likewise, E-plate negative from the positive plates to travel, But losing potential energy and kinetic energy increase, so, E = KE_i + PE_i = KE_f + PE_f. You KE_f (= (1 / 2) * I have enough information to calculate Is given * v_f ^ 2) And you know, and PE_i PE_f, you I already know, so you can easily find and v_i KE_i was calculated.

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