How much charge is on each plate of a 3.00-μF capacitor when it is connected toa 15.0-V battery? b) If this same capacitor is c
1 answer:
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Answer:
<em>20.08 Volts</em>
Explanation:
<u>Parallel Connection of Capacitors</u>
The voltage across any two elements connected in parallel is the same. If the elements are capacitors, then each voltage is
They are both the same after connecting them, thus
Or, equivalently
The total charge of both capacitors is
We can compute the total charge by using the initial conditions where both capacitors were disconnected:
Now we compute Q1 from the equation above
The final voltage of any of the capacitors is
Answer:
The change in potential energy is
Explanation:
From the question we are told that
The magnitude of the uniform electric field is
The distance traveled by the electron is
Generally the force on this electron is mathematically represented as
Where F is the force and q is the charge on the electron which is a constant value of
Thus
Generally the work energy theorem can be mathematically represented as
Where W is the workdone on the electron by the Electric field and is the change in kinetic energy
Also workdone on the electron can also be represented as
Where considering that the movement of the electron is along the x-axis
So
substituting values
Now From the law of energy conservation
Where is the change in potential energy
Thus