This problem involves Coulomb's Law, which describes the electrostatic force between charged particles. The force depends on the magnitude of the charges, the distance between them, and the medium in which they are placed. The dielectric constant of a medium quantifies how the medium affects the electric field and, consequently, the force between charges.
Coulomb's Law in Vacuum: The force between two charges \(q_1\) and \(q_2\) separated by a distance \(r\) in a vacuum is given by:
\[ F_{\text{vacuum}} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \]Here, \(\epsilon_0\) is the permittivity of free space. The problem states that this initial force is \(F\).
\[ F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \]Coulomb's Law in a Dielectric Medium: When the charges are placed in a medium with a dielectric constant \(\kappa\) (also denoted as \(K\) or \(\epsilon_r\)), the permittivity of the medium is \(\epsilon = \epsilon_0 \kappa\). The force in this medium, \(F_{\text{medium}}\), is given by:
\[ F_{\text{medium}} = \frac{1}{4\pi\epsilon} \frac{q_1 q_2}{r^2} = \frac{1}{4\pi\epsilon_0 \kappa} \frac{q_1 q_2}{r^2} \]Relating the Forces: We can rewrite \(F_{\text{medium}}\) in terms of \(F_{\text{vacuum}}\) (which is \(F\)):
\[ F_{\text{medium}} = \frac{1}{\kappa} \left( \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \right) \] \[ F_{\text{medium}} = \frac{F_{\text{vacuum}}}{\kappa} \]Given \(F_{\text{vacuum}} = F\) and the dielectric constant \(\kappa = 6\), the new force \(F_{\text{new}}\) is:
\[ F_{\text{new}} = \frac{F}{6} \]Correct Option: C) The new force is \(F/6\). This is because the electrostatic force between charges in a dielectric medium is reduced by a factor equal to the dielectric constant of the medium compared to the force in a vacuum.
This problem requires the application of Coulomb's Law, which describes the electrostatic force between two point charges. We are given the magnitudes of the charges, the distance between them, and the medium (vacuum). We need to calculate the force using the formula for Coulomb's Law.
The electrostatic force \(F\) between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) in vacuum is given by:
\[ F = k \frac{|q_1 q_2|}{r^2} \]In vacuum, the value of the electrostatic constant \(k\) (also known as Coulomb's constant) is approximately:
\[ k = 9 \times 10^9 \text{ N m}^2/\text{C}^2 \]Now, plug the given values into Coulomb's Law equation:
\[ F = (9 \times 10^9 \text{ N m}^2/\text{C}^2) \frac{|(1 \text{ C})(1 \text{ C})|}{(1 \text{ m})^2} \]C) 9 × 10⁹ N is the correct answer. This is the calculated electrostatic force between two 1 Coulomb charges separated by 1 meter in a vacuum, a very large force illustrating why 1 Coulomb is a very large unit of charge.
The question asks about the conditions under which Coulomb's law is valid. Coulomb's law describes the electrostatic force between charged particles. Understanding its fundamental assumptions is key to answering this question.
A) Point charges at rest
This option perfectly matches the conditions under which Coulomb's law is defined and applicable. It applies to charges that are considered point-like and are not in motion relative to each other.
The question asks for the charge on an electron. This is a fundamental constant in physics and chemistry. We need to recall its standard value and sign.
Correct Option: C) -1.6 × 10⁻¹⁹ C
This option correctly states both the negative sign and the magnitude of the elementary charge, which is the charge carried by an electron.
The question asks about the meaning of "quantization of electric charge". This is a fundamental concept in physics, specifically in electrostatics. We need to identify the option that correctly defines this principle.
D) Charge exists in integral multiples of e
This statement perfectly describes the principle of quantization of electric charge. It means that the total charge on any object is always an integer multiple of the elementary charge \(e\).
Two identical point charges repel each other with a force of 100 N when placed 0.5 m apart. The magnitude of each charge is approximately:
5 × 10⁻⁴ C
1.67 × 10⁻³ C
5.27 × 10⁻⁵ C
1.67 × 10⁻⁴ C
This problem can be solved using Coulomb's Law, which describes the electrostatic force between two point charges. We are given the force, the distance between the charges, and the fact that the charges are identical. We need to find the magnitude of each charge.
C) \(5.27 \times 10^{-5} \text{ C}\) is the calculated magnitude of each charge, which matches the option.