/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Two spheres charged with \(100 \... [FREE SOLUTION] | 91Ó°ÊÓ

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Two spheres charged with \(100 \mu \mathrm{C}\) and \(-100 \mu \mathrm{C}\) are kept at a distance. The force acting on them is \(F_{1}\). They are connected with a metallic wire and then conductor is removed. The force \(F_{2}\) acting on them now will be : (a) equal to \(F_{1}\) (b) more than \(F_{1}\) (c) zero (d) infinite

Short Answer

Expert verified
The force \(F_2\) acting on them is zero. (Option c)

Step by step solution

01

Understand Initial Conditions

Initially, the two spheres have charges of \(100 \, \mu C\) and \(-100 \, \mu C\). The force \(F_1\) acting on them is due to the electrostatic force between these opposite charges, as calculated by Coulomb's law.
02

Connect Spheres with a Metallic Wire

When the spheres are connected by a metallic wire, charges flow between them aiming to equalize the potential. Total charge is \(Q_\text{total} = 100 \, \mu C + (-100 \, \mu C) = 0\). Hence, both spheres end up uncharged.
03

Remove the Wire and Recalculate Force

After the wire is removed, since both spheres are uncharged, the electrostatic force between them becomes zero. According to Coulomb's law, if either charge is zero, then the force must also be zero.

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Key Concepts

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

Electrostatic Force
Electrostatic force is a fundamental force that acts between charges. It is described by Coulomb's Law, which calculates the force between two point charges. If you have charges like the ones from the exercise, where one charge is positive and the other is negative, the electrostatic force between them can be attractive.

Using Coulomb's Law, the force magnitude is given by: \( F = k \frac{|q_1q_2|}{r^2} \) where:
  • \(F\) is the force between charges,
  • \(k\) is Coulomb's constant \(8.99 \times 10^9 \mathrm{N \cdot m^2/C^2}\),
  • \(q_1\) and \(q_2\) are the amounts of the charges,
  • \(r\) is the distance between the centers of the charges.
With opposite charges like the ones in the exercise, the force \(F_1\) is attractive. Understanding this concept is crucial, because it explains why charged objects can exert forces over a distance.
Charge Neutralization
Charge neutralization happens when two charged bodies come to the point where the total charge is balanced or zero. In the exercise, this is demonstrated when the two spheres, initially having opposite charges of \(100 \, \mu \text{C}\) and \(-100 \, \mu \text{C}\), are connected by a metallic wire.

The connection allows the charges to balance out since charges will flow to equalize potential. The total charge becomes zero because:\[ Q_\text{total} = 100 \, \mu \text{C} + (-100 \, \mu \text{C}) = 0 \]This leads to both spheres being uncharged after removing the wire. Consequently, the electrostatic force, which depends on the presence of charges, becomes nonexistent, which is why the force \(F_2\) equals zero after charge neutralization has occurred.
Metallic Conductor Interaction
When a metallic conductor is introduced between charged objects, it allows for the free flow of charges. Metals are good conductors due to a sea of free electrons that can move easily. This property is utilized in the exercise when the spheres are connected by a metallic wire.

The interaction ensures that the charges redistribute themselves until there is no potential difference between the spheres. Essentially, this metallic conductor acts as a bridge for charge flow, achieving equilibrium. This is how charge neutralization occurs, as outlined in the previous section.

Once the wire is removed, it doesn't reintroduce charges; it merely ensures the charges have had a chance to disperse and equalize. This lack of residual charge explains why, once the wire is disconnected, the spheres exert no electrostatic force on each other, resulting in \(F_2\) being zero.

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Most popular questions from this chapter

The intensity of an electric field inside a capacitor is \(E\). The work required to make a charge \(q\) move in a closed rectangular circuit is : (a) \(2(l+b) q E\) (b) \(2 l q E\) (c) \(2 b g E\) (d) zero

The capacitance (C) for an isolated conducting sphere of radius \((a)\) is given by \(4 \pi \varepsilon_{0} a .\) If the sphere is enclosed with an earthed concentric sphere, the ratio of the radii of the spheres being \(\frac{n}{(n-1)}\) then the capacitance of such a sphere will be increased by a factor? (a) \(n\) (b) \(\frac{n}{(n-1)}\) (c) \(\frac{(n-1)}{n}\) (d) \(a n\)

A capacitor of capacitance \(C\) is charged to a potential difference \(V_{0}\). The charged battery is disconnected and the capacitor is connected to a capacitor of unknown capacitance \(C_{x}\). The potential difference across the combination is \(V\). The value of \(C_{r}\) should be : (a) \(\frac{C\left(V_{0}-V\right)}{V}\) (b) \(\frac{C\left(V-V_{0}\right)}{V}\) (c) \(\frac{\mathrm{CV}}{V_{0}}\) (d) \(\frac{C V_{0}}{V}\)

Two condensers of capacities \(2 C\) and \(C\) are joined in parallel and charged upto potential \(V .\) The battery is removed and the condenser of capacity \(C\) is filled completely with a medium of dielectric constant \(K\). The potential difference across the capacitors will now be: (a) \(\frac{3 V}{K+2}\) (b) \(\frac{3 V}{K}\) (c) \(\frac{V}{K+2}\) (d) \(\frac{V}{K}\)

A sheet of aluminium foil of negligible thickness is placed between the plates of a capacitor of capacitance \(C\) as shown in the figure then capacitance of capacitor becomes: (a) \(2 \mathrm{C}\) (b) \(\mathrm{C}\) (c) C/2 (d) zero

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