TBR physics electricity circuits Phase II passage III

[deleted783484]

Hi can someone help me with this question please!

It's a passage but I will type out the part that we need:

they gave us a circuit with a power supply- a resistor R1 and a capacitor with a solution of cell suspended in a solution known as a dielectric constant: they are talking about electroporation

If the electric field is strong enough then the cells undergo dielectric breakdown opening micropores in the cell membranes. dielectric breakdown seperates the charges of a dipole momentarily so that ions fill in the space in between the plates of the capacitor. The capacitor now discharges through the cells and the cells act as resistors to the current flow. During discharge there is likely problem that cell lysis can occur. The rate of lysing can be expressed as:

L= 1-(t/tc)exp(-(E-Ec)/K)
where tc= treshold value of treatment time
t= actual treatment time
Ec= treshold Efield strength
E= actual Efield strength
K= constant


QUESTION: CELL LYSIS IS AFFECTED BY WHICH OF THE FOLLOWING FACTORS
I. AN INCREASE IN CAPACITOR PLATE AREA
II. A DECREASE IN PLATE SEPERATION
III. INCREASING THE VALUE OF R1
IV. THE NUMBER OF CELLS BEING TREATED

ANSWER is all I,II,IV

I picked all of them and don't understand why R1 doesn't affect discharge. TBR explanation is charging R1 would affect only how long it takes the capacitor to charge up. it has no effect on discharging across the capacitor and the passage states that discharging the capacitor is what sends a current through the cells that may cause them to lyse

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[BerkReviewTeach]

The cells are in the region of space between the plates of the capacitor, so they are affected by the charges accruing on those plates. Once the plates are fully charged, no current is running through the single-loop circuit, so R1 is irrelevant at that point. The only role R1 plays is hindering the flow of current and thereby causing the plates to take a longer time to reach full charge.

The discharge of current is from one plate to the other, through the gel and not through the rest of the circuit. The flow of charge through the collapsing field is only affected by the contents between the plates, although if the capacitor is still hooked up, then there is the chance that current can flow through the circuit in attempt to reestablish a voltage across the capacitor. But that charge is through the wires and will not be the ones affecting the cells.

The point of this question is that the cells are in the capacitor, so anything affecting the capacitance and voltage across the capacitor, will influence the lysing.

 

[deleted783484]

The cells are in the region of space between the plates of the capacitor, so they are affected by the charges accruing on those plates. Once the plates are fully charged, no current is running through the single-loop circuit, so R1 is irrelevant at that point. The only role R1 plays is hindering the flow of current and thereby causing the plates to take a longer time to reach full charge.

The discharge of current is from one plate to the other, through the gel and not through the rest of the circuit. The flow of charge through the collapsing field is only affected by the contents between the plates, although if the capacitor is still hooked up, then there is the chance that current can flow through the circuit in attempt to reestablish a voltage across the capacitor. But that charge is through the wires and will not be the ones affecting the cells.

The point of this question is that the cells are in the capacitor, so anything affecting the capacitance and voltage across the capacitor, will influence the lysing.

Thank you for your reply! What do you mean by no current is running through the single-loop circuit? I thought that no current would run through the capacitor but current would still run through the resistors when the capacitor is charging?

 

[BerkReviewTeach]

When the capacitor is charging, there is current through the entire circuit (there has to be to charge the plates). But once fully charged, the capacitor repels charge flow from the voltage source, causing charge flow through the circuit to cease. As long as the capacitor is fully charged, that circuit will have no current.