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Chapter 17: Problem 10

There are two charged identical metal spheres \(A\) and \(B\) repel each other with a force \(3 \times 10^{-5} \mathrm{~N}\). Another identical uncharged sphere \(C\) is touched with \(A\) and then placed at the mid-point between \(A\) and \(B\). Net force on \(C\) is (a) \(1 \times 10^{-5} \mathrm{~N}\) (b) \(2 \times 10^{-5} \mathrm{~N}\) (c) \(1.5 \times 10^{-5} \mathrm{~N}\) (d) \(3 \times 10^{-5} \mathrm{~N}\)

Short Answer

Expert verified
(c) \(1.5 \times 10^{-5} \mathrm{~N}\)

Step by step solution

01

Understanding Initial Conditions

Initially, spheres \(A\) and \(B\) repel each other with a force of \(3 \times 10^{-5} \text{ N}\). This indicates that the spheres have equal but opposite charges, \( q_A \) and \( q_B \). Because \(A\) and \(B\) are identical and repel each other, we have that \(q_A = q_B \).
02

Charging Sphere C

Sphere \(C\) is initially uncharged. When \(C\) is touched to sphere \(A\), charge will redistribute equally because they are identical metal spheres, each taking half of the total charge. After touching, each sphere \(A\) and \(C\) will have a charge of \( \frac{q_A}{2} \). Since initially \(A\) had charge \( q_A \), after touching, both \(A\) and \(C\) now have \( \frac{q_A}{2} \).
03

Calculating New Forces on Sphere C

After sphere \(C\) is placed at the midpoint between \(A\) and \(B\), let's calculate the forces. Sphere \(A\) exerts a force \(F_{AC}\) on \(C\) given by Coulomb鈥檚 law: \[ F_{AC} = k \cdot \frac{\left(\frac{q_A}{2}\right) \left( \frac{q_A}{2} \right)}{(0.5d)^2} \] Similarly, sphere \(B\) exerts a force \(F_{BC}\) on \(C\) because \(B\)'s charge remains \(q_B = q_A\): \[ F_{BC} = k \cdot \frac{q_A \cdot \left( \frac{q_A}{2} \right)}{(0.5d)^2} \] Because \(A\) and \(B\) are equidistant from \(C\) and since \(F_{BC} > F_{AC}\), net force \(F_C = \lvert F_{BC} - F_{AC} \lvert = \frac{1}{2} F_{AB}\).
04

Calculating Net Force

Plug the given force value \( F_{AB} = 3 \times 10^{-5} \text{ N} \) into \(F_C = \frac{1}{2} F_{AB}\): \[ F_C = \frac{1}{2} \times \left(3 \times 10^{-5} \text{ N}\right) = 1.5 \times 10^{-5} \text{ N}\] So, the net force on sphere \(C\) is \(1.5 \times 10^{-5} \text{ N}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coulomb's Law
Coulomb's Law is fundamental in understanding how charged objects interact. It defines the force of attraction or repulsion between two charged bodies. This force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The formula is expressed as: \[ F = k \cdot \frac{q_1 \cdot q_2}{r^2} \]where:
  • F is the force between the charges.
  • k is Coulomb's constant \( (8.9875 \times 10^9 \text{ Nm}^2/\text{C}^2) \).
  • q_1 and q_2 are the magnitudes of the charges.
  • r is the distance between the centers of the two charges.
This law helps predict how charges will behave in various situations. Such as the initial problem statement where two metal spheres, each carrying charge, repel due to the like charges exerting a force on one another.
Charge Distribution
The concept of charge distribution becomes essential when analyzing the touching of sphere A with sphere C. When a charged sphere touches an identical uncharged sphere, the charge distributes evenly. This is because the spheres are conductive and share equal potential. Initially, sphere A had charge \(q_A\). Upon touching sphere C, the charge equally divides, giving both spheres a charge of \(\frac{q_A}{2}\).
  • After touching, both have \(\frac{q_A}{2}\) each because they are identical.
  • Equal distribution occurs due to identical size and material.
Charge distribution is a crucial concept in electrostatics. It illustrates how proximity affects electrical equilibrium and prepares us for further force calculations in such scenarios.
Force Calculation
Force calculation is important to determine the net force acting on an object in an electric field. After sphere C was charged and placed at the midpoint between spheres A and B, we used Coulomb鈥檚 Law to find individual forces.The force of sphere A on C \((F_{AC})\) and sphere B on C \((F_{BC})\) are calculated using:\[ F = k \cdot \frac{q \cdot q}{d^2} \]Given the charges and position:
  • A provides a reduced repelling force (half the charge), so \(F_{AC}\) is less.
  • B, being at full charge, allows \(F_{BC}\) to be stronger.
The net force is the difference \(F_C = |F_{BC} - F_{AC} |\). Using the formula:\[ F_C = \frac{1}{2} F_{AB} = 1.5 \times 10^{-5} \text{ N} \]This calculation effectively demonstrates how proximity and charge magnitude impact force magnitude.

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Most popular questions from this chapter

Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximately. \(\quad\) [NCERT Exemplar] (a) spheres (b) planes (c) paraboloids (d) ellipsoids

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The electric field in the space between the plates of a discharge tube is \(3.25 \times 10^{-4} \mathrm{NC}^{-1}\). If mass of proton is \(1.67 \times 10^{-27} \mathrm{~kg}\) and its charge is \(1.6 \times 10^{-19} \mathrm{C}\), the force often the proton in the field is (a) \(10.4 \times 10^{-15} \mathrm{~N}\) (b) \(20 \times 10^{-23} \mathrm{~N}\) (c) \(5.40 \times 10^{-15} \mathrm{~N}\) (d) \(5.20 \times 10^{-15} \mathrm{~N}\)

The force on each plate of parallel plate capacitor has a magnitude equal to \(\frac{1}{2} Q E\), where \(Q\) is the charge on the capacitor and \(E\) is the magnitude of electric field between the plates. Then (a) \(\frac{E}{2}\) contributes to the force against which the plates are moved (b) \(\frac{E}{3}\) contributes to the force against which the plates are moved (c) \(E\) contributes force against which the plates are moved (d) None of the above

The dimensional forumula of absolute permittivity of air or free space \(\left(\varepsilon_{0}\right)\) is (a) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]\) (b) \(\left[\mathrm{M}^{\mathrm{D}} \mathrm{L}^{-3} \mathrm{~T}^{3} \mathrm{~A}^{3}\right]\) (c) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{-3} \mathrm{~A}\right]\) (d) \(\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{TA}^{2}\right]\)
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