Question

A 10 μC charge is divided into two equal parts and kept at 1 cm distance. Find repulsion between charges?

• A. \( 7 \, \mu \text{C}, 3 \, \mu \text{C} \)
• B. \( 8 \, \mu \text{C}, 2 \, \mu \text{C} \)
• C. \( 5 \, \mu \text{C}, 5 \, \mu \text{C} \)
• D. \( 9 \, \mu \text{C}, 1 \, \mu \text{C} \)

Answer 1

The Correct Option is C

To solve the problem of finding the repulsion between two charges resulting from a 10 μC charge divided into two equal parts and kept at a 1 cm distance, we will use Coulomb's Law. Coulomb's Law provides a way to calculate the electrostatic force between two point charges. The formula is given by:

\(F = \frac{k \cdot |q_1 \cdot q_2|}{r^2}\)

where:

• \( F \) = electrostatic force between the charges
• \( k \) = Coulomb's constant, approximately \( 8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)
• \( q_1 \) and \( q_2 \) = magnitudes of the charges
• \( r \) = distance between the charges

 

Let the total charge \( q = 10 \, \mu \text{C} \) be divided into two parts, \( x \) and \( (q - x) \). The force between the two charges is given by Coulomb's law: \[ F = \frac{k \, x \, (q - x)}{r^2}, \] where \( k \) is Coulomb's constant and \( r = 1 \, \text{cm} \) is the distance between the charges. To maximize \( F \), differentiate \( F \) with respect to \( x \) and set the derivative equal to zero: \[ \frac{dF}{dx} = 0. \] Simplify: \[ \frac{d}{dx} \left( \frac{k \, x \, (q - x)}{r^2} \right) = \frac{k}{r^2} \left( q - 2x \right) = 0. \] This gives: \[ q - 2x = 0. \] Solve for \( x \): \[ x = \frac{q}{2} = \frac{10 \, \mu \text{C}}{2} = 5 \, \mu \text{C}. \] The other part of the charge is: \[ q - x = 10 \, \mu \text{C} - 5 \, \mu \text{C} = 5 \, \mu \text{C}. \] Thus, the charges of the two parts are \( \boxed{5 \, \mu \text{C}} \) and \( \boxed{5 \, \mu \text{C}} \).