Two charges of equal magnitudes and at a distance r exert a force F on each other. If the charges are halved and distance between them is doubled, then the new force acting on each charge is

To find the new force acting on each charge after halving their magnitudes and doubling the distance between them, we can use Coulomb's Law. This law states that the force \( F \) 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 them. The formula is given by:

Coulomb's Law

The formula can be expressed as:

F = k * (|q1 * q2|) / r²

where:

• F is the force between the charges,
• k is Coulomb's constant,
• q1 and q2 are the magnitudes of the charges,
• r is the distance between the charges.

Initial Conditions

Let's denote the initial charges as \( q \) and \( q \) (since they are equal), and the initial distance as \( r \). According to Coulomb's Law, the initial force \( F \) can be expressed as:

F = k * (q * q) / r²

New Conditions

Now, if we halve the charges, each charge becomes \( \frac{q}{2} \). The new distance is doubled, so it becomes \( 2r \). We can substitute these new values into Coulomb's Law to find the new force \( F' \):

F' = k * \left(\frac{q}{2} * \frac{q}{2}\right) / (2r)²

Calculating the New Force

Let's simplify this expression step by step:

• The product of the charges becomes \( \frac{q^2}{4} \) (since \( \frac{q}{2} * \frac{q}{2} = \frac{q^2}{4} \)).
• The square of the new distance is \( (2r)² = 4r² \).

Now substituting these into the formula gives:

F' = k * \left(\frac{q^2}{4}\right) / (4r²)

This simplifies to:

F' = k * \frac{q^2}{16r²}

Relating the New Force to the Initial Force

Recall that the initial force was:

F = k * \frac{q^2}{r²}

Now, we can express \( F' \) in terms of \( F \):

F' = \frac{1}{16} * \left(k * \frac{q^2}{r²}\right) = \frac{1}{16} * F

Final Result

Thus, the new force acting on each charge after halving the charges and doubling the distance is:

F' = \frac{F}{16}

This means the force is reduced to one-sixteenth of the original force. This example illustrates how both the magnitude of the charges and the distance between them significantly influence the electrostatic force acting between them.