A 5.7 F capacitor and a 250 Ω resistor are connected to a battery of voltage 6 V as shown in the circuit. After closing the switch, while charging the capacitor, when the capacitor voltage is half of the battery voltage, how much charge is stored on each plate of the capacitor? Express your answer in Coulombs (C).

When the capacitor voltage is half of the battery voltage, we can use the formula Q = CV, where Q is the charge stored in the capacitor, C is the capacitance, and V is the voltage across the capacitor.

First, let's calculate the charge stored in the capacitor when the voltage is half of the battery voltage. The battery voltage is 6 V, so when the capacitor voltage is half of that, it is 3 V.

Using the formula [tex]Q = CV,[/tex] we can substitute the values:

Q = (5.7 F)(3 V)
Q = 17.1 C

So, when the capacitor voltage is half of the battery voltage, there is a charge of 17.1 Coulombs stored in each plate of the capacitor.

(Note: The given values and calculations in this explanation are for illustrative purposes and may not reflect the actual values in the problem.)

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