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Figure 3 shows an arrangement used to investigate the repulsive forces between two identical charged conducting spheres - AQA - A-Level Physics - Question 4 - 2019 - Paper 2

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Figure 3 shows an arrangement used to investigate the repulsive forces between two identical charged conducting spheres. The spheres are suspended by non-conducting ... show full transcript

Worked Solution & Example Answer:Figure 3 shows an arrangement used to investigate the repulsive forces between two identical charged conducting spheres - AQA - A-Level Physics - Question 4 - 2019 - Paper 2

Step 1

Calculate the potential of one of the spheres.

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Answer

To find the potential (V) of a sphere, we can use the formula:

V=QCV = \frac{Q}{C}

where:

  • Q = charge on the sphere = 52 nC = 52 × 10^-9 C
  • C = capacitance of the sphere = 4πεr

Given r = 20 mm = 0.020 m, we first calculate the capacitance:

C=4π(8.85×1012)(0.020)5.58×1012 FC = 4\pi(8.85 \times 10^{-12})(0.020) \approx 5.58 \times 10^{-12} \text{ F}

Now substituting into the potential formula:

V=52×1095.58×10129330.5VV = \frac{52 \times 10^{-9}}{5.58 \times 10^{-12}} \approx 9330.5 \, \text{V} Thus, the potential of one of the spheres is approximately 9330.5 V.

Step 2

Draw labelled arrows on Figure 3 to show the forces on sphere B.

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On sphere B, there are three forces to consider:

  1. The electrostatic repulsive force (F_e) acting away from sphere A.
  2. The weight (W) of sphere B acting downwards due to gravity.
  3. The tension (T) in the thread acting upwards along the line of the thread.

Arrows should clearly indicate these directions, with appropriate labels.

Step 3

Suggest a solution to one problem involved in the measurement of d in Figure 3.

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One problem in measuring the distance d is the difficulty in ensuring that the spheres do not influence each other's positions during the measurement. A possible solution is to use a non-conductive ruler or laser distance measuring device to accurately gauge the distance without affecting the charges or the gravitational interactions of the spheres.

Step 4

Show that the magnitude of the electrostatic force on each sphere is about 4 × 10^-3 N.

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Answer

The electrostatic force between two charged spheres can be calculated using Coulomb's law:

F=kQ1Q2r2F = k \frac{Q_1 Q_2}{r^2}

where:

  • k = 8.99 × 10^9 N·m²/C²
  • Q_1 and Q_2 = 52 nC = 52 × 10^{-9} C
  • r = the distance between the centers of the spheres = d = 40 mm = 0.040 m.

Substituting the values:

F=8.99×109(52×109)(52×109)(0.040)24.05×103NF = 8.99 \times 10^9 \frac{(52 \times 10^{-9})(52 \times 10^{-9})}{(0.040)^2} \approx 4.05 \times 10^{-3} N Thus, the magnitude of the electrostatic force is approximately 4 × 10^{-3} N.

Step 5

Discuss whether this measurement is consistent with the other data in this investigation.

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Answer

The angle θ measured by the student is 7°. Since the system is in equilibrium, this angle should align with the forces acting on the sphere due to tension, weight, and electrostatic forces. By applying trigonometric relationships, we can show that an angle of 7° corresponds to the calculated forces, thus supporting that the measurement is consistent with other data.

Step 6

Deduce with a calculation whether this statement is valid.

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Answer

To assess the significance of gravitational forces, we consider the weight of each sphere:

W=mg=(3.2×103kg)(9.81m/s2)0.0314NW = mg = (3.2 \times 10^{-3} kg)(9.81 m/s^2) \approx 0.0314 \, \text{N}

Comparing the gravitational force and the electrostatic force:

  • Fe4×103NF_e \approx 4 × 10^{-3} N
  • W0.0314NW \approx 0.0314 N

Since WW is significantly larger than FeF_e, it can be deduced that the gravitational force does have a notable effect on the system, thus contradicting the student's statement.

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