Question 109: Electrostatics & Capacitance

Question

The force between two electric point charges at rest in air is $$F_1$$ When the same arrangement is kept inside water, the force between them is $$F_2$$. Which of the following statement is correct?

Accepted Answer

The force between two point charges is described by Coulomb's Law, which states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between their centers. In a vacuum or air, this force can be given by the equation: $$F = k\frac{|q_1 q_2|}{r^2}$$ where: $F$ is the magnitude of the force between the charges, $q_1$ and $q_2$ are the magnitudes of the charges, $r$ is the distance between the charges, $k$ is Coulomb's constant (approximately $8.987 imes 10^9 \, ext{N m}^2/ ext{C}^2$ in vacuum or air). When the same arrangement of charges is placed in a medium such as water, the force between the charges is affected by the dielectric constant ($\varepsilon_r$) of the medium. The equation for the force in a medium becomes: $$F' = \frac{k}{\varepsilon_r}\frac{|q_1 q_2|}{r^2}$$ where $\varepsilon_r$ is the relative permittivity (dielectric constant) of the medium. For water, $\varepsilon_r$ is approximately 80 at room temperature. Because $\varepsilon_r > 1$ for any medium other than vacuum, the effect of this is to reduce the force between the charges: $$F_2 = \frac{F_1}{\varepsilon_r}$$ Given that $\varepsilon_r > 1$ for water, this implies that $\frac{1}{\varepsilon_r} < 1$, hence $F_2 < F_1$. Therefore, the correct statement is: Option D: $F_2 < F_1$

Answer

$$F_2=0$$

Answer

$$F_2>F_1$$

Answer

$$F_2=F_1$$

Answer

$$F_2< F_1$$

Electrostatics & Capacitance

The force between two electric point charges at rest in air is $F_1 When the same arrangement is kept inside water, the force between them is F_2$. Which of the following statement is correct?

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