Increasing resistance and the effect on voltage

[dr2453]

Homework Statement

Which of the following will most likely increase the electric field between the plates of a parallel plate capacitor?
(a) Adding a resistor that is connected to the capacitor in series
(b) Adding a resistor that is connected to the capacitor in parallel
(c) Increasing the distance between the plates
(d) Adding an extra battery to the system

Homework Equations

E=V/d
V=iR

The Attempt at a Solution

(c) was automatically out, as pe r E=V/d. I am stuck between between a, b, and c. I know that adding a battery would increase V, thus increasing E. However, I'm confused about whether increasing resistance decreases or increases voltage. According to V=iR, it seems like it should (there's nothing in the problem indicating whether i is constant). Does increasing resistance decrease or increase the voltage applied to the capacitor?

 

[dr2453]

sorry, (b) was supposed to be Adding a resistor that is connected to the capacitor in parallel.

 

[domenico]

(d) mentions a battery, so we are talking about dc current. In this case, the circuit is open at the capacitor. So, adding a resistor in series doesn't change anything, this eliminates (a). Neither does adding a resistor in parallel, because the voltage stays the same. Since you ruled out (c) by yourself, the correct answer would be (d).

 

[dr2453]

Thanks! That's what I thought the answer would be. But why does adding a resistor not change anything?

 

[BruceW]

Since it is d.c. the current through the capacitor is in one direction only. So what will happen to the charge on the capacitor as time increases? And after a long enough time has passed, what will happen then?

 

[domenico]

@dr2453, the voltage between battery terminals is constant, so current cannot flow across the capacitor. This is because the capacitor's plates are separated by a dielectric film or air, only a changing electric field (voltage) would make the current pass through it.
So, if a resistor it's added in series, no current would flow through it, and the capacitor would be directly connected to the battery, as if the resistor wasn't there. The same goes if the resistor is added in parallel, the capacitor is still directly connected to the battery and nothing changes.
@BruceW, I'm sure this problem isn't about transitory response, so there's no need to involve time. However, in that case the charge will increase exponentially with time, until it reaches the value Q= C*V, where C is the capacitance and V is the voltage of the battery.