# Current flow in a capacitor circuit on insertion of dielectric [closed] Approach 1)Since a dielectric alters only the charge on the plates, i.e. The charge decreases as the dielectric moves inside the capacitor and then finally after reaching the middle position charge on plate increases.

2)Therefore current must flow from the positive end to the negative one i.e.A to B.

Are the above arguments correct?

given

CASE 1 The dielectric slab is moved through the capacitor t=0 completely outside at t=t1 completely inside and at some time t=t2 it has passed through the capacitor

CASE 2 the dielectric is simply inserted between the plates

Disclaimer not a homework question, a question from a recent exam.

• My guess is that current always flows clockwise. Insertion of dielectric will cause the capacitance to increase and thus to accumulate more charge onto the plates, current will flow clockwise. – Apoorv Potnis Mar 27 '17 at 11:40
• I read this as the dielectric first enters the region between the plates and then exits (is passed between the plates). – Alfred Centauri Mar 27 '17 at 11:47
• A dielectric does not alter the charge on the plate; rather, it decreases the electric field (and decrease the potential difference across the capacitor). The electric field between the plates decreases due to polarization. Check this answer to understand how polarization works. – Yashas Mar 27 '17 at 11:49
• so do u mean the capacitance increases and hence the current increases initially reaches max value then decreases – Sher Mar 27 '17 at 11:55
• The current reverses direction as the dielectric passes between the plates. – Farcher Mar 27 '17 at 15:29

I believe your reasoning is partially correct, there is just one point that is unclear.

You can assume that the insulating plate is a better dielectric than air. Inserting the plate, then, would increase the capacitance (the insulator polarizes i.e. polar molecules align with the field, and effectively reduces the electric force between the plates with their own contribution of electric force in the opposite direction). Since capacitance increases, the ability to store charge increases (it takes more charge to cause the same voltage across the plates of the capacitor since the electric force/electric field is weaker and the "voltage" is the electric field with distance taken out of account) and so more charge will flow onto the plates.

Given all that, a conventional current will flow from positive to negative - from A to B, just like you said. What seems slightly ambiguous to me is whether or not the insulator passes all the way through or if it is simply inserted in between the plates. If it is simply inserted, current will only flow from A to B. But if it passes straight through, then the capacitance will decrease after the half way point and extra charge difference between the plates will cause a stronger electric force/electric field, thus a higher voltage between the plates, than was originally supplied. Charge will then flow in the opposite direction - from B to A.

Note: I usually prefer to talk in electric force in my answers as opposed to electric field. This helps me remember that the "field" that everyone talks about is just the electric force, without taking the charge it acts on into account. Electric field is used because you don't know what the charge that feels a force will be, but it "muddies the waters" when talking like this because electric field doesn't sound like electric force.

• given that the slab moves through the capacitor i.e. at t=0 completely outside at t=t1 completely inside and at some time t=t2 it has passed through the capacitor – Sher Mar 27 '17 at 12:48
• The potential difference across the capacitor does not change.$A$ has a positive charge and $B$ has a negative charge before the dielectric is inserted.You quite rightly state that as the dielectric is inserted the capacitance increases and so more charge is stored on the plates, $Q\uparrow = C\uparrow V$.There must be more positive charge on plate $A$ and more negative charge (less positive charge) on plate $B$.The (conventional - movement of positive charges) current must be from $B$ along the wires and through the cell to $A$. The current will reverse as the dielectric leaves the capacitor. – Farcher Mar 27 '17 at 13:20

Insertion of a dielectric does not directly affect the charge on the plates (unless you insert it slowly); rather, it decreases the electric field between the plates. As the electric field and the potential are related by $V = Ed$, the potential difference across the capacitor also decreases ($d$ remains constant; $E$ decreases, therefore $V$ decreases).

Case 1: Capacitor is fully charged before insertion

Initially, the potential difference across the capacitor would be equal to the E.M.F of the battery. As the insertion of dielectric reduces the potential difference across the capacitor, a current is driving in the circuit such that the potential drop across the capacitor equals the E.M.F. If the potential drop across the capacitor has to increase, the charge on the plates must increase; therefore, the current flows in the direction which increases charge on the capacitor. This direction is from $A$ to $B$.

Case 2: Capacitor is not fully charged; on its way to being fully charged

Initially, the potential difference across the capacitor is less than the E.M.F of the battery. When the dielectric is inserted, the potential difference drops. For the reasons mentioned in the previous case, the current will flow in such a way that the charge on the capacitor increases. This direction is from $A$ to $B$.

Case 3: Capacitor is overcharged; potential drop across the capacitor is more than the E.M.F

Initially, the potential difference across the capacitor is more than the E.M.F. This would cause the current to charge the battery (initial current is from $B$ to $A$). After the dielectric is inserted, the potential drop across the capacitor decreases. There are two sub-cases in this situation.

• The new potential drop is smaller than the E.M.F

The current reverses. The current flows such that the potential drop across the capacitor increases (tries to fully charge the capacitor). The direction of current is from $A$ to $B$.

• The new potential drop is greater than the E.M.F

As the potential drop across the capacitor is greater than the E.M.F, the current continues to flow from $B$ to $A$.

• the question says the dielectric is moved through the capacitor! i.e. at t=0 completely outside at t=t1 completely inside and at some time t=t2 it has passed through the capacitor – Sher Mar 27 '17 at 12:50
• Aw, then my answer does not answer your question but it does tell you how to proceed. The capacitance goes to a max and then decreases. The answer is both $(3)$ and $(4)$ depending on the case. – Yashas Mar 27 '17 at 12:51
• still thanks for it quite great to know don't delete it let me add this part in the question so your answer remains here – Sher Mar 27 '17 at 12:52
• Conventional current direction as the dielectric is inserted into the capacitor is from $B$ to $A$. See my comment to the other answer. – Farcher Mar 27 '17 at 13:26
• @Farcher shouldn't the direction depend on the velocity at which the dielectric is inserted? – Yashas Mar 27 '17 at 13:55

Assume that the potential difference across the capacitor does not change; it stays equal to the emf of the cell.
Basically I am assuming the resistance of the wires and cell are relatively small and so is the time constant of the circuit.
$A$ has a positive charge and $B$ has a negative charge before the dielectric is inserted.
As the dielectric is inserted the capacitance increases and so more charge is stored on the plates.
$C\uparrow V \Rightarrow Q\uparrow$
The amount of positive charge on plate $A$ must increase and there will also be more negative charge (less positive charge) on plate $B$.
The (conventional - movement of positive charges) current must flow from $B$ along the wires and through the cell to $A$.

The current will reverse as the dielectric leaves the capacitor; its capacitance decreasing and the amount of charge stored being less.