The method and the results are on the electric field lab. Do the analysis yourself first, then check the answers below.
Quick answer: An electric field E = F/q is the force per unit positive charge, a vector drawn with field lines. These answers go with the AP Physics 2 lab on electric field: the graph, the calculations, the uncertainties and the conclusion explained, for all 7 analysis questions.
The sample readings give a gradient of 60 V/m, the same as V/d. If your data differ slightly, check that the line is straight and passes through the origin, and give the unit of every result.
Level: AP Physics 2 (algebra-based) · Curriculum: AP P2 Unit 10
The points lie on a straight line through the origin, so the potential rises at a constant rate with distance. V is directly proportional to x, which means the field is uniform.
Question 2 Gradient and field
Gradient = (6.00 − 0.00) / (10.0 − 0) = 0.600 V/cm = 60.0 V/m. Therefore E = 60 V/m, directed from the positive to the negative electrode.
Question 3 Compare with V/d
V/d = 6.0 / 0.100 = 60 V/m. The gradient is also 60.0 V/m, so the difference is 0 %, within the precision of the voltmeter.
Question 4 Energy of a charge
W = qΔV = 1.0 × 10⁻⁶ × 6.0 = 6.0 × 10⁻⁶ J.
Question 5 Equipotentials and field lines
Equipotentials are straight lines parallel to the electrodes. Field lines are drawn at right angles to them, parallel to each other and pointing from high to low potential.
Question 6 Voltmeter resistance
A high-resistance voltmeter draws almost no current from the paper, so it does not disturb the potential it measures. A low-resistance meter would lower the reading and distort the pattern. High resistance, negligible current.
Question 7 Edge effects
At the edges the current can spread outward and the field lines bulge, so lines curve. Use a wider sheet, or take readings only along the central region. Take readings in the centre.
