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Capacitors

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Question 9

A student is using a spreadsheet to model the decay of charge on a capacitor.

They are using the equation

ΔQΔt=−Q4.0 \frac{\Delta Q}{\Delta t} = -\frac{Q}{4.0} ΔtΔQ​=−4.0Q​

The student chooses a time interval of 1.0 s. At time t=0.0 st = 0.0\text{ s}t=0.0 s the charge on the capacitor is 800 μC800\text{ }\mu\text{C}800 μC.

Part of the modelling spreadsheet is shown below.

t / st\text{ / s}t / sCharge Q Q\,Q left on capacitor after time t / μCt\text{ / }\mu\text{C}t / μCCharge ΔQ \Delta Q\,ΔQ decaying in the next 1.0 s / μC1.0\text{ s / }\mu\text{C}1.0 s / μC
0.0800200
1.0600
2.0
3.0
4.0

What is the charge on the capacitor at t=3.0 st = 3.0\text{ s}t=3.0 s?

200 μC200\text{ }\mu\text{C}200 μC

338 μC338\text{ }\mu\text{C}338 μC

378 μC378\text{ }\mu\text{C}378 μC

450 μC450\text{ }\mu\text{C}450 μC

Capacitors Questions

  1. A Level
  2. /Physics
  3. /Capacitors