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Date November 2019 Marks available 2 Reference code 19N.2.HL.TZ0.9
Level Higher level Paper Paper 2 Time zone 0 - no time zone
Command term Calculate Question number 9 Adapted from N/A

Question

X has a capacitance of 18 μF. X is charged so that the one plate has a charge of 48 μC. X is then connected to an uncharged capacitor Y and a resistor via an open switch S.

The capacitance of Y is 12 μF. S is now closed.

Calculate, in J, the energy stored in X with the switch S open.

[2]
a.

Calculate the final charge on X and the final charge on Y.

[3]
b(i).

Calculate the final total energy, in J, stored in X and Y.

[2]
b(ii).

Suggest why the answers to (a) and (b)(ii) are different.

[2]
c.

Markscheme

E=12Q2C OR V=QC

E=«1248×10-618×10-6=» 6.4×10-5«J»

a.

ALTERNATIVE 1
QX+QY=48 ✔

QX18=QY12

solving to get QX=29«μC»  QY=19«μC»

 

ALTERNATIVE 2

48=18V+12VV=1.6«V»

QX=«1.6×18=» 29 «QX=1.6×18=29«μC» ✔

QY=«1.6×12=»19«μC» ✔

 

NOTE: Award [3] for bald correct answer

b(i).

ALTERNATIVE 1

ET=1229×10-6218×10-6+1219×10-6212×10-6

=3.8×10-5«J»

 

ALTERNATIVE 2

ET=12×18×10-6×1.62+12×12×10-6×1.62 

=3.8×10-5«J»

 

NOTE: Allow ECF from (b)(i)
Award [2] for bald correct answer
Award [1] max as ECF to a calculation using only one charge

b(ii).

charge moves/current flows «in the circuit» ✔
thermal losses «in the resistor and connecting wires» ✔

NOTE: Accept heat losses for MP2

c.

Examiners report

[N/A]
a.
[N/A]
b(i).
[N/A]
b(ii).
[N/A]
c.

Syllabus sections

Additional higher level (AHL) » Topic 11: Electromagnetic induction » 11.3 – Capacitance
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Additional higher level (AHL) » Topic 11: Electromagnetic induction
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