Electrostatic Numericals: Relative Permittivity of Medium, free space & Force

Problem 1: Force in Free Space

Two charges of \(1\,\mu\text{C}\) (i.e. \(1 \times 10^{-6}\,\text{C}\)) are placed \(3\,\text{cm}\) apart in free space. Using Coulomb’s law, calculate the repulsive force between them. (Take \(k = 8.99 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2\).)

Select the correct option:

• A) \(10\,\text{N}\)
• B) \(9\,\text{N}\)
• C) \(11\,\text{N}\)
• D) \(12\,\text{N}\)

Detailed Explanation

Coulomb's law in free space is given by: \[ F = k \frac{q^2}{r^2}. \]

Here, \(q = 1 \times 10^{-6}\,\text{C}\), \(r = 3\,\text{cm} = 0.03\,\text{m}\), and \(k = 9 \times 10^9\).

Compute: \[ F = 9 \times 10^9 \times \frac{(1 \times 10^{-6})^2}{(0.03)^2} = 9 \times 10^9 \times \frac{1 \times 10^{-12}}{9 \times 10^{-4}} \approx 10\,\text{N}. \]

Hence, the correct answer is Option A.

Problem 2: Force in a Dielectric Medium

Two charges of \(2\,\mu\text{C}\) (i.e. \(2 \times 10^{-6}\,\text{C}\)) are placed \(4\,\text{cm}\) apart in oil with a dielectric constant \(\kappa = 5\). Calculate the repulsive force between the charges. (Use \(k = 8.99 \times 10^9\,\text{N}\cdot\text{m}^2/\text{C}^2\).)

Select the correct option:

• A) \(4.5\,\text{N}\)
• B) \(2.25\,\text{N}\)
• C) \(9.0\,\text{N}\)
• D) \(5.0\,\text{N}\)

Detailed Explanation

When a dielectric is present, the effective Coulomb constant becomes \(k/\kappa\). Thus: \[ F = \frac{k}{\kappa} \frac{q^2}{r^2}. \]

Here, \(q = 2 \times 10^{-6}\,\text{C}\), \(r = 4\,\text{cm} = 0.04\,\text{m}\), \(\kappa = 5\), and \(k = 9 \times 10^9\).

Calculate the modified constant: \[ \frac{k}{\kappa} = \frac{9 \times 10^9}{5} \approx 1.8 \times 10^9. \]

Now compute: \[ F = 1.8 \times 10^9 \times \frac{(2 \times 10^{-6})^2}{(0.04)^2} = 1.8 \times 10^9 \times \frac{4 \times 10^{-12}}{1.6 \times 10^{-3}} \approx 4.5\,\text{N}. \]

Therefore, the correct answer is Option A.

Problem 3: Force Reduction Due to Dielectric

In free space, the force between two charges is measured to be \(5 \times 10^{-3}\,\text{N}\). When the space between them is completely filled with a dielectric medium having a constant \(\kappa = 2.0\), what is the new force between the charges?

Select the correct option:

• A) \(5 \times 10^{-3}\,\text{N}\)
• B) \(2.5 \times 10^{-3}\,\text{N}\)
• C) \(1 \times 10^{-2}\,\text{N}\)
• D) \(1.25 \times 10^{-3}\,\text{N}\)

Detailed Explanation

In a dielectric medium the force reduces by a factor equal to \(\kappa\): \[ F' = \frac{F}{\kappa}. \]

Given \(F = 5 \times 10^{-3}\,\text{N}\) and \(\kappa = 2.0\), \[ F' = \frac{5 \times 10^{-3}}{2.0} = 2.5 \times 10^{-3}\,\text{N}. \]

Hence, the correct answer is Option B.

Problem 4: Determining the Dielectric Constant

Two charges in air repel each other with a force of \(8 \times 10^{-4}\,\text{N}\). When a dielectric medium is inserted between them, the force reduces to \(2 \times 10^{-4}\,\text{N}\). Find the dielectric constant \(\kappa\) of the medium.

Select the correct option:

• A) 2
• B) 4
• C) 0.5
• D) 8

Detailed Explanation

Inserting a dielectric reduces the force by the factor \(\kappa\): \[ F' = \frac{F}{\kappa}. \]

Rearranging gives: \[ \kappa = \frac{F}{F'}. \]

Here, \(F = 8 \times 10^{-4}\,\text{N}\) and \(F' = 2 \times 10^{-4}\,\text{N}\), so: \[ \kappa = \frac{8 \times 10^{-4}}{2 \times 10^{-4}} = 4. \]

Thus, the correct answer is Option B.

Problem 5: Force Calculation in a Dielectric Medium

Given that the relative permittivity of free space is 1 and that of a certain medium is 6, two charges of \(3\,\mu\text{C}\) (i.e. \(3 \times 10^{-6}\,\text{C}\)) are placed \(5\,\text{cm}\) apart in this medium. Calculate the repulsive force between the charges.

Select the correct option:

• A) \(5.4\,\text{N}\)
• B) \(6.8\,\text{N}\)
• C) \(4.2\,\text{N}\)
• D) \(7.1\,\text{N}\)

Detailed Explanation

In a medium with dielectric constant \(\kappa\), Coulomb's law reads: \[ F = \frac{k}{\kappa} \frac{q^2}{r^2}. \]

Here, \(q = 3 \times 10^{-6}\,\text{C}\), \(r = 5\,\text{cm} = 0.05\,\text{m}\), \(\kappa = 6\), and \(k = 9 \times 10^9\).

First, compute the effective constant: \[ \frac{k}{6} \approx 1.5 \times 10^9. \]

Then, calculate: \[ F = 1.5 \times 10^9 \times \frac{(3 \times 10^{-6})^2}{(0.05)^2} = 1.5 \times 10^9 \times \frac{9 \times 10^{-12}}{2.5 \times 10^{-3}}. \]

Numerator: \(1.5 \times 10^9 \times 9 \times 10^{-12} \approx 1.35 \times 10^{-2}\).
Denominator: \(2.5 \times 10^{-3}\).
Thus, \[ F \approx \frac{1.35 \times 10^{-2}}{2.5 \times 10^{-3}} \approx 5.4\,\text{N}. \]

Therefore, the correct answer is Option A.