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Edexcel | Physics WPH0 | W 11 6 1 | 1 hour 15 minutes
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  • Q1
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3
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A student measures the energy stored in a capacitor of unknown capacitance.

She charges the capacitor to a potential difference V using a battery and then discharges

the capacitor through a joulemeter which records the energy W stored in the capacitor.

She uses two different batteries and records the following readings.

3
(a) (I)
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2 Mark s

For each potential difference, calculate the mean energy W stored in the

capacitor. Hence calculate the capacitance C using the formula W=½CV 2.

Add your values to the table.

Mean W values are 19.57 and 36.16 

Values for C are 1.9(3) and 2.0(1) to 2 or 3 sf

3
(a) (II)
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1 Mark

Calculate the percentage difference between your two values of C.

Percentage difference =

Percentage difference between values calculated

using mean C as denominator 

e.g. (2.01 – 1.93)/1.97 = 4(.1)%

3
(b)
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The uncertainty in the values of potential difference in the table is 0.1 V.

3
(b) (I)
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1 Mark

Estimate the uncertainty in your mean value of W when using the 4.5 V battery.

Uncertainty =

Uses range or half range to estimate uncertainty 

[must include unit]

e.g uncertainty is 0.12 mJ or 0.06 mJ

3
(b) (II)
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2 Mark s

Use these uncertainties to estimate the percentage uncertainty in the value of C

obtained using the 4.5 V battery.

Percentage uncertainty =

Calculates percentage uncertainties in V and W 

Combines percentages appropriately 

e.g. using range (0.12 mJ) using half range (0.06 mJ)

W = 0.6% & V = 2.2% W = 0.3% & V = 2.2%

C = [0.6+(2×2.2)]%=5.0% C = [0.3+(2×2.2)]%=4.7%

3
(c)
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2 Mark s

tolerance of 20%

Calculates % difference between their mean C and 2200 μF (1)

Compares % difference (+ % uncertainty) with 20% (1)

OR

Calculates lower limit of range of 2200 μF as1760 (1)

States that that mean C( – uncertainty) lies within range (1)

e.g %difference from quoted value is (2.2 – 1.95)/2.2 = 11%

%D + %U = 16 % which is within tolerance of 20%

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