Consider the arrangement of point charges in the figure.
a) Determine the direction of the force on q (with q > 0) in the figure, given that qa = qb = +7.50 µC and qc = qd = -7.50 µC.
b) Calculate the magnitude of the force on the charge q in newtons, given that the square is 7 cm on a side and q = 2.75 µC.
The Correct Answer and Explanation is:
To solve this problem, we consider a square configuration of point charges with a central charge q. The charges are arranged such that:
- qa = qb = +7.50 µC (positive charges)
- qc = qd = -7.50 µC (negative charges)
- Charge q = +2.75 µC is located at the center of the square
- The square has sides of 7.00 cm = 0.07 m
a) Direction of the force on q (q > 0)
Each corner charge exerts an electrostatic force on the central charge q. Because opposite charges attract and like charges repel, we analyze the net force vectorially:
- qa (+) and qb (+) (top-left and top-right corners, for example) will repel the central positive charge q, pushing it diagonally away from them.
- qc (-) and qd (-) (bottom-left and bottom-right corners) will attract the central positive charge q, pulling it toward them.
By symmetry:
- The horizontal components of the forces from qa and qb cancel each other.
- The horizontal components of the forces from qc and qd also cancel.
- The vertical components from qa and qb both point downward (repelling q),
- The vertical components from qc and qd also point downward (attracting q).
Thus, all vertical components add, resulting in a net force directed straight downward.
✅ Answer for part (a): The net force on q is directed straight downward.
b) Magnitude of the force on q
Each charge at a corner exerts a force on q. Since q is at the center, the distance from each corner to the center is:
$$
r = \frac{\text{diagonal}}{2} = \frac{\sqrt{2} \cdot 0.07}{2} \approx 0.0495 \text{ m}
$$
Using Coulomb’s law for the magnitude of the force between two point charges:
$$
F = \frac{k |q_1 q_2|}{r^2}
$$
Where:
- $k = 8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2$
- $q_1 = 7.50 \times 10^{-6} \, \text{C}$
- $q_2 = 2.75 \times 10^{-6} \, \text{C}$
- $r = 0.0495 \, \text{m}$
$$
F_{single} = \frac{(8.99 \times 10^9)(7.5 \times 10^{-6})(2.75 \times 10^{-6})}{(0.0495)^2} \approx 7.56 \, \text{N}
$$
Each force contributes only its vertical component:
$$
F_{vertical_single} = F_{single} \cdot \sin(45^\circ) = 7.56 \cdot \frac{\sqrt{2}}{2} \approx 5.35 \, \text{N}
$$
There are four charges, each contributing to a downward component:
$$
F_{net} = 4 \cdot 5.35 = \boxed{21.4 \, \text{N downward}}
$$
✅ Answer for part (b): The magnitude of the net force on q is approximately 21.4 N, directed downward.
Summary
- The symmetrical placement of charges leads to cancellation of horizontal components.
- Vertical components reinforce due to all pushing/pulling in the same direction.
- The total force is directed downward with a magnitude of approximately 21.4 N.
