Capacitor Diagram: Investigating Charging Process

[Masafi]

I have attached the document which has the question and the diagram

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A student is learning about how capacitors work. He uses the circuit shown in the attacment to investigate the capacitor C.

Letter X labels a connection which he can make to either of the points L or M. Each cell has an e.m.f. of 1.5 V.

He connects X to L. He sketches how the reading on ammeter 1 varies with time (Figure 2).

Explain in terms of charge what has happened in the circuit.

The answer is:

1. Charging process (1)
2. Plates oppositely charged OR charge moves from one plate to another (1)
3. Charge flows anticlockwise OR electrons flow clockwise OR left
4. plate becomes positive OR right plate becomes negative (1)
5. Build up of Q/V reduces flow rate (1)

I don't understand the last 2 points, can someone explain them please. How do we determine the direction of charge? And what do they mean with the capacitance building up.

 

[Stonebridge]

Point 4 should read
Left plate becomes positive OR right plate becomes negative
Point 5 is badly presented. It (Q/V) doesn't mean Q divided by V reduces flow, it means build up of Q OR build up of V reduces flow.
The polarity of the cell determines the direction of the flow of charge when charging.

 

[Masafi]

Point 4 should read
Left plate becomes positive OR right plate becomes negative
Point 5 is badly presented. It (Q/V) doesn't mean Q divided by V reduces flow, it means build up of Q OR build up of V reduces flow.
The polarity of the cell determines the direction of the flow of charge when charging.

Thanks.

4. Was a typo by me sorry. So why does the left plate become positive? Is it because the corresponding left side of cell (the line to the right of the resistor) is positive?

5. Oh ok. Why is that the build up of Q or V reduces flow? I = Q/t, so if Q increases, I increases?

 

[Stonebridge]

4. The cell pushes electrons clockwise round the circuit onto the right plate of the capacitor. The negative charge on the right plate repels negative charge off the left plate making it positively charged through loss of electrons.
5. The initial flow of current from the battery in the circuit starts to charge up the capacitor. As more charge flows on to the capacitor its pd rises. The pd on the capacitor is in the opposite direction to the battery, and opposes it. As the pd builds up on the capacitor there is more opposition to current and the current gets smaller. Eventually the pd on the capacitor is equal (but opposite) to that of the battery and no more current flows. The capacitor is then "fully" charged.

 

[Masafi]

4. The cell pushes electrons clockwise round the circuit onto the right plate of the capacitor. The negative charge on the right plate repels negative charge off the left plate making it positively charged through loss of electrons.

How do u know that it goes clockwise?

5. The initial flow of current from the battery in the circuit starts to charge up the capacitor. As more charge flows on to the capacitor its pd rises. The pd on the capacitor is in the opposite direction to the battery, and opposes it. As the pd builds up on the capacitor there is more opposition to current and the current gets smaller. Eventually the pd on the capacitor is equal (but opposite) to that of the battery and no more current flows. The capacitor is then "fully" charged.

So the pd on the capacitor is always against the pd of the battery? Why is that?

 

[Stonebridge]

How do u know that it goes clockwise?



So the pd on the capacitor is always against the pd of the battery? Why is that?

1. Check the diagram. Which side of the battery (left or right) is negative? Which way does negative charge flow from the negative terminal of the battery?
2. Because negative charges repel other negative charges. If you push negative charges onto something (the plate of a capacitor), they will repel any more charges wanting to get to the same place.