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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

The potential ( V ) of a sphere can be calculated using the formula:

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

where ( Q ) is the charge on each sphere (52 nC = ( 52 \times 10^{-9} , \text{C} )) and ( C ) is the capacitance of the sphere given by ( C = 4 \pi \epsilon_{0} r ).

The radius ( r ) is 20 mm = 0.020 m. Plugging in the values:

C=4π(8.85×1012)(0.020)2.22×1012FC = 4 \pi (8.85 \times 10^{-12}) (0.020) \approx 2.22 \times 10^{-12} \, \text{F}

Thus,

V=52×1092.22×101223,000VV = \frac{52 \times 10^{-9}}{2.22 \times 10^{-12}} \approx 23,000 \, \text{V}

Therefore, the potential of one of the spheres is approximately 23,000 V.

Step 2

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

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Answer

Label the forces acting on sphere B as follows:

  1. An upward arrow to represent the tension in the thread.
  2. A downward arrow indicating the gravitational force acting on sphere B.
  3. A horizontal arrow to the left showing the electrostatic repulsive force due to sphere A.

Step 3

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

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Answer

One problem in measuring the distance d accurately is the inability to observe the exact point at which the centers of the spheres are aligned. To address this, a ruler with a finer scale could be used to measure d, ensuring that it is positioned parallel to the line connecting the centers 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

Using Coulomb's law, the electrostatic force ( F ) between two charged spheres is given by:

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

where ( k \approx 8.99 \times 10^9 , \text{N m}^2/ ext{C}^2 ), ( Q_1 ) and ( Q_2 ) are the charges on the spheres (both 52 nC), and ( r ) is the distance between the centers of the spheres (d = 0.040 m).

Substituting the values:

F=8.99×109×(52×109)2(0.040)24×103NF = 8.99 \times 10^9 \times \frac{(52 \times 10^{-9})^2}{(0.040)^2} \approx 4 \times 10^{-3} \, \text{N}

Step 5

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

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Answer

The angle of 7° observed by the student can be analyzed in context with the equilibrium of forces acting on the spheres. If the calculated electrostatic force (4 × 10⁻³ N) is significantly greater than the gravitational force acting on the spheres, this will contribute to a larger angle at equilibrium. The consistent results would suggest the angle is reasonable provided that the gravitational force is minimal compared to the electrostatic force.

Step 6

Deduce with a calculation whether this statement is valid.

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Answer

To assess the validity of the student’s statement regarding gravitational force, calculate the gravitational force:

Fg=mgF_g = mg

Where:

  • Mass ( m = 3.2 \times 10^{-3} , \text{kg} )
  • Acceleration due to gravity ( g = 9.81 , \text{m/s}^2 )

Thus,

Fg=(3.2×103)(9.81)0.0314NF_g = (3.2 \times 10^{-3})(9.81) \approx 0.0314 \, \text{N}

Comparing with the earlier computed electrostatic force ( F \approx 4 \times 10^{-3} , \text{N} ):

The gravitational force is approximately 0.0314 N, which is relatively small compared to the electrostatic force; hence, the statement made by the student is not valid as the gravitational force does affect the equilibrium state.

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