Capacitors lab answers for AP Physics 2 (free PDF download)

The method and the results are on the capacitors lab. Do the analysis yourself first, then check the answers below.

Quick answer: A capacitor stores charge on two plates separated by an insulator, with capacitance C = Q/V in farads. These answers go with the AP Physics 2 lab on capacitors: the graph, the calculations, the uncertainties and the conclusion explained, for all 7 analysis questions.

Capacitors lab answers for AP Physics 2 (free PDF download)

The sample data give a gradient of 3.54 × 10⁻¹³ F·m and ε₀ near 8.85 × 10⁻¹² F/m. If your values are different, check that your graph is straight and that you converted units, and state how large the stray capacitance was.

Level: AP Physics 2 (algebra-based) · Curriculum: AP P2 Unit 10

Contents

Question 1 Calculate 1/d

d = 1.0, 2.0, 3.0, 4.0 and 5.0 mm gives 1/d = 1000, 500, 333, 250 and 200 m⁻¹.

Question 2 Graph and gradient

Gradient = (362 − 79) pF / (1000 − 200) m⁻¹ = 0.354 pF·m = 3.54 × 10⁻¹³ F·m. The graph is straight, so C is proportional to 1/d plus a constant.

Question 3 Value of ε₀

Gradient = ε₀A, so ε₀ = 3.54 × 10⁻¹³ / 0.040 = 8.85 × 10⁻¹² F/m, in line with the accepted value.

Question 4 The intercept

The line is C = 0.354 × (1/d) + c, and at 1/d = 200 we have 79 = 70.8 + c, so c ≈ 8 pF. This is stray capacitance from the leads and the meter.

Question 5 Dielectric constant

κ = (410 − 8) / (126 − 8) = 402 / 118 = 3.4.

Question 6 Uncertainty in d

1.0 mm: 0.05/1.0 = 5 %. 5.0 mm: 0.05/5.0 = 1 %. Small gaps give the largest uncertainty, so errors affect the left of the graph most.

Question 7 Prediction

At 1/d = 100 m⁻¹, C = 0.354 × 100 + 8 = 43 pF. If the sheets bend, the real gap varies and the effective d is smaller than measured, so C reads high.

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