electricity - shocking! Flashcards
(15 cards)
potential energy of charges
PE=-qEdeltaX
PE= QV
Why does this work?
Well, E= V/x, so in essence
PE=qV/x*x or simply
PE=qV
Note also that qE=F (electric force equals charge times electric field, the go between equation) SO really the first equation is no more than force*distance (work)
Voltage
kq/r
charge per distance
V=Q/Capacitance
Joules/coulomb
Work per charge
To increase voltage means to increase the amount of work that each charge can do.
Also, V=electric field * distance
kq/d^2*d= kq/d
Does it take energy to move a charge along an equipotential line? why or why not?
PE = QdeltaV
If the voltage doesn’t change, then no work is done
Capacitors
C=Q/V
coulomb/volt
charge per (charge per distance or work per charge)
epsilonA/d
Explain why a dielectric increases the maximum operating voltage of a capacitor even though the physical size of the capacitor doesn’t change.
The material of the dielectric may be able to withstand a larger electric field than air can withstand before breaking down to pass a spark between the capacitor plates.
When a dielectric is introduced, does the voltage go up or down?
Down
Being said, the battery’s voltage does not change. Does this mean that the voltage between the plates can actually be greater than the battery?
What is the potential energy of a capacitor?
PE=1/2QV
When you add a dielectric, what do you do??
C=KCnaught
C=KepsilonA/d
You can increase capacitance by increasing area, decreasing distance.
C=Q/V
You can also increase capacitance by decreasing the voltage (like using a dielectric).
When you are wondering about capacitance and electric fields and the possibility that the dielectric might break down, what do you remember?
C=Q/V
and V=electric field times distance
Often you’re not going to be given the voltage, but you’ll prob be given a distance and a max field strength before breakdown. There ya go.
capacitors in series
capacitance is the product/sum
charges same
I=I1=I2=I3=…
voltages add
V=V1+V2+V3+…
capacitors in parallel
capacitance simply adds
charges simply add
voltage is the same
resistors in series
resistance simply adds
current same
voltage adds
resistors in parallel
resistance is product/sum
different current
same voltage
power?
P=IV
plate of charges
E=Q/2epsilonA
parallel plate
E=Q/epsilonA
bc the positive negative plates reinforce each other