Consider the circuit below. If the battery voltage is doubled, what happens to t

Question

Consider the circuit below. If the battery voltage is doubled, what happens to the electric field in the resistor? What effect would this here on the to the electrical force on electrons traveling through the resistor? Explain. The circuit below is used to measure an unknown capacitance C in series with a resistor with R = 10 M Ohm. The switch is initially in position A for a long time. At time t = 0, the switch is moved to position B. The voltage across the capacitor at t = 60 seconds later is measured to be 2V. What is the capacitance of the capacitor?

Explanation / Answer

the electric field defind as

E = V/L

when V i sdoubled E also doubles

force acting on the electrons is

F = Eq

when V is doubled E also doubled which leads twice force acting on electrons

F '= 2Eq= 2 F

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