Answer :
Certainly! Let's solve the given problem step by step.
1. Understanding the initial conditions:
- Initially, the force between the two opposite charges is [tex]\( F \)[/tex].
- The initial force is given as [tex]\( \frac{16}{25} F \)[/tex].
2. Charge transfer:
- We are transferring [tex]\( 60\% \)[/tex] of the charge from one charge to another.
- Let's denote the initial charges as [tex]\( q \)[/tex] and [tex]\( -q \)[/tex], where [tex]\( q \)[/tex] and [tex]\( -q \)[/tex] are the magnitudes of the charges.
3. New charges after transfer:
- After transferring [tex]\( 60\% \)[/tex] of the charge from one to another, the new charges will be as follows:
- The new charge of the first object [tex]\( q_1' \)[/tex] will be [tex]\( q - 0.60q = 0.40q \)[/tex].
- The new charge of the second object [tex]\( q_2' \)[/tex] will be [tex]\( -q + 0.60q = -0.40q \)[/tex].
4. Calculating the new force:
- The force between two point charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] separated by a distance [tex]\( r \)[/tex] is given by Coulomb's law:
[tex]\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
- In this context, we assume the distance [tex]\( r \)[/tex] remains the same and thus constant.
- The new force [tex]\( F' \)[/tex] between the charges [tex]\( q_1' \)[/tex] and [tex]\( q_2' \)[/tex] can be calculated in a proportional manner since the proportionality constant [tex]\( k \)[/tex] and [tex]\( r \)[/tex] are not changing.
Therefore, for the new charges:
[tex]\[ F' \propto (0.40q) \cdot (-0.40q) = -0.16q^2 \][/tex]
- We notice that the sign on the force indicates attraction because the charges are opposite. However, the magnitude of the force we primarily care about is [tex]\( 0.16q^2 \)[/tex].
5. Relating the new force to the initial force:
- The initial force was proportional to [tex]\( q \cdot -q = -q^2 \)[/tex].
- Therefore, the initial force [tex]\( F \)[/tex] is:
[tex]\[ F = k \frac{q^2}{r^2} \][/tex]
- We establish a proportion between the initial and the new force:
[tex]\[ \frac{F'}{F} = \frac{-0.16q^2}{-q^2} = 0.16 \][/tex]
- Hence, the new force [tex]\( F' \)[/tex] is:
[tex]\[ F' = \frac{16}{25} F \cdot 0.16 = \left(\frac{16}{25} \cdot 0.16\right) F = \frac{16 \times 0.16}{25} F = \frac{16 \times 16}{25 \times 100}F = \frac{256}{2500} F = 0.1024 F \][/tex]
6. Numerical Result:
- Based on accurate calculations done earlier, we now denote:
[tex]\[ F' = -0.16000000000000003 \][/tex]
- The negative sign denotes attraction but we consider the magnitude for the final answer.
Therefore, after transferring [tex]\( 60\% \)[/tex] of the charge from one charge to another, the new force between the two charges is approximately [tex]\(-0.16000000000000003 \approx -0.16\)[/tex].
Thus, the new force is [tex]\( \approx -0.16F \)[/tex].
1. Understanding the initial conditions:
- Initially, the force between the two opposite charges is [tex]\( F \)[/tex].
- The initial force is given as [tex]\( \frac{16}{25} F \)[/tex].
2. Charge transfer:
- We are transferring [tex]\( 60\% \)[/tex] of the charge from one charge to another.
- Let's denote the initial charges as [tex]\( q \)[/tex] and [tex]\( -q \)[/tex], where [tex]\( q \)[/tex] and [tex]\( -q \)[/tex] are the magnitudes of the charges.
3. New charges after transfer:
- After transferring [tex]\( 60\% \)[/tex] of the charge from one to another, the new charges will be as follows:
- The new charge of the first object [tex]\( q_1' \)[/tex] will be [tex]\( q - 0.60q = 0.40q \)[/tex].
- The new charge of the second object [tex]\( q_2' \)[/tex] will be [tex]\( -q + 0.60q = -0.40q \)[/tex].
4. Calculating the new force:
- The force between two point charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] separated by a distance [tex]\( r \)[/tex] is given by Coulomb's law:
[tex]\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \][/tex]
- In this context, we assume the distance [tex]\( r \)[/tex] remains the same and thus constant.
- The new force [tex]\( F' \)[/tex] between the charges [tex]\( q_1' \)[/tex] and [tex]\( q_2' \)[/tex] can be calculated in a proportional manner since the proportionality constant [tex]\( k \)[/tex] and [tex]\( r \)[/tex] are not changing.
Therefore, for the new charges:
[tex]\[ F' \propto (0.40q) \cdot (-0.40q) = -0.16q^2 \][/tex]
- We notice that the sign on the force indicates attraction because the charges are opposite. However, the magnitude of the force we primarily care about is [tex]\( 0.16q^2 \)[/tex].
5. Relating the new force to the initial force:
- The initial force was proportional to [tex]\( q \cdot -q = -q^2 \)[/tex].
- Therefore, the initial force [tex]\( F \)[/tex] is:
[tex]\[ F = k \frac{q^2}{r^2} \][/tex]
- We establish a proportion between the initial and the new force:
[tex]\[ \frac{F'}{F} = \frac{-0.16q^2}{-q^2} = 0.16 \][/tex]
- Hence, the new force [tex]\( F' \)[/tex] is:
[tex]\[ F' = \frac{16}{25} F \cdot 0.16 = \left(\frac{16}{25} \cdot 0.16\right) F = \frac{16 \times 0.16}{25} F = \frac{16 \times 16}{25 \times 100}F = \frac{256}{2500} F = 0.1024 F \][/tex]
6. Numerical Result:
- Based on accurate calculations done earlier, we now denote:
[tex]\[ F' = -0.16000000000000003 \][/tex]
- The negative sign denotes attraction but we consider the magnitude for the final answer.
Therefore, after transferring [tex]\( 60\% \)[/tex] of the charge from one charge to another, the new force between the two charges is approximately [tex]\(-0.16000000000000003 \approx -0.16\)[/tex].
Thus, the new force is [tex]\( \approx -0.16F \)[/tex].
