To solve this problem, we need to analyze the forces acting on the charge \( Q \) as it moves in the electric field created by the two fixed charges \( A \) and \( B \). Let's break it down step by step.
Understanding the Setup
We have two fixed charges, \( A \) and \( B \), each with a charge of \( 5 \, \mu C \) (microcoulombs), separated by a distance of \( 6 \, m \). The midpoint \( C \) is located \( 3 \, m \) from both \( A \) and \( B \). A charge \( Q \) of \( -5 \, \mu C \) is shot from point \( C \) perpendicular to the line joining \( A \) and \( B \) with an initial kinetic energy of \( 0.06 \, J \).
Electric Field Calculation
First, we need to calculate the electric field \( E \) at point \( C \) due to charges \( A \) and \( B \). The electric field \( E \) created by a point charge is given by the formula:
E = k \cdot \frac{|Q|}{r^2}
where \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, N \cdot m^2/C^2 \)), \( Q \) is the charge, and \( r \) is the distance from the charge to the point where the field is being calculated.
Calculating the Electric Field at Point C
Since both charges are equal and located symmetrically about point \( C \), the electric fields due to \( A \) and \( B \) at point \( C \) will have the same magnitude but opposite directions along the line joining them. The distance from each charge to point \( C \) is \( 3 \, m \).
- For charge \( A \):
EA = \( k \cdot \frac{5 \times 10^{-6}}{(3)^2} = 8.99 \times 10^9 \cdot \frac{5 \times 10^{-6}}{9} \approx 4.99 \times 10^3 \, N/C \
- For charge \( B \):
EB = \( k \cdot \frac{5 \times 10^{-6}}{(3)^2} = 4.99 \times 10^3 \, N/C \
Since both electric fields point away from the positive charges, the net electric field \( E \) at point \( C \) is:
E = EA + EB = 4.99 \times 10^3 + 4.99 \times 10^3 = 9.98 \times 10^3 \, N/C
Force on Charge Q
The force \( F \) acting on charge \( Q \) due to the electric field \( E \) is given by:
F = Q \cdot E
Substituting the values:
F = \( -5 \times 10^{-6} \cdot 9.98 \times 10^3 \approx -4.99 \times 10^{-2} \, N \
The negative sign indicates that the force is attractive, pulling charge \( Q \) towards the line joining \( A \) and \( B \).
Energy Considerations
As charge \( Q \) moves from point \( C \) to point \( D \), it converts its kinetic energy into electric potential energy. The initial kinetic energy \( KE \) is given as \( 0.06 \, J \). The potential energy \( U \) at point \( D \) can be expressed as:
U = k \cdot \frac{|Q_A \cdot Q|}{r}
As charge \( Q \) comes to rest at point \( D \), we have:
KE + U = 0
Thus, we can set up the equation:
0.06 = k \cdot \frac{(5 \times 10^{-6})(-5 \times 10^{-6})}{r}
Rearranging gives us:
r = k \cdot \frac{(5 \times 10^{-6})(-5 \times 10^{-6})}{0.06}
Calculating Distance CD
Substituting \( k \) and simplifying:
r = \( \frac{(8.99 \times 10^9)(25 \times 10^{-12})}{0.06} \approx 3.74 \, m \
Since \( C \) is the midpoint, the distance \( CD \) is approximately \( 3.74 \, m \). Therefore, the charge \( Q \) comes to rest at a distance of about \( 3.74 \, m \) from point \( C \).
Final Thoughts
This problem illustrates the interplay between kinetic energy and electric forces. By understanding how electric fields work and how they affect charged particles, we can predict the behavior of charges in various configurations. If you have any further questions or need clarification on any steps, feel free to ask!
