In the figure four particles form a square. The charges are q1 = +Q, q2 = q3 = q, and q4 = -21.0Q. What is q/Q if the net electrostatic force on particle 1 is zero?
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The Correct Answer and Explanation is:
To find the ratio qQ\frac{q}{Q} such that the net electrostatic force on particle 1 is zero, we must analyze the forces acting on it due to the other three charges arranged in a square.
Step-by-step Setup
Given:
- Four charges form a square of side length aa.
- q1=+Qq_1 = +Q (Top-left corner)
- q2=qq_2 = q (Top-right corner)
- q3=qq_3 = q (Bottom-left corner)
- q4=−21Qq_4 = -21Q (Bottom-right corner)
We need to find qQ\frac{q}{Q} such that the net force on charge q1q_1 is zero.
Step 1: Understand the Geometry
Label the square:
- Let’s place the square in the x-y plane with side length aa.
- q1q_1 is at (0, a)
- q2q_2 is at (a, a)
- q3q_3 is at (0, 0)
- q4q_4 is at (a, 0)
Step 2: Calculate Forces on q1q_1
Force due to q2=qq_2 = q:
- Distance: aa
- Direction: Left (−x)
- Magnitude: F12=kQqa2F_{12} = k \frac{Qq}{a^2}
Force due to q3=qq_3 = q:
- Distance: aa
- Direction: Down (−y)
- Magnitude: F13=kQqa2F_{13} = k \frac{Qq}{a^2}
Force due to q4=−21Qq_4 = -21Q:
- Distance: 2a\sqrt{2}a
- Direction: Along the diagonal (from (a,0) to (0,a))
- The force is attractive, pointing toward q4q_4
- Components of the force:
- F14=kQ⋅21Q(2a2)=21kQ22a2F_{14} = k \frac{Q \cdot 21Q}{(2a^2)} = \frac{21kQ^2}{2a^2}
- Break into x and y: both components are equal since the diagonal makes a 45° angle.
- So: F14x=F14y=21kQ22a2⋅cos(45∘)=21kQ22a2⋅12=21kQ222a2F_{14x} = F_{14y} = \frac{21kQ^2}{2a^2} \cdot \cos(45^\circ) = \frac{21kQ^2}{2a^2} \cdot \frac{1}{\sqrt{2}} = \frac{21kQ^2}{2\sqrt{2}a^2}
Step 3: Set Net Force to Zero
We want net force on q1q_1 to be zero in both x and y directions.
X-direction:
- Leftward force from q2q_2: F12x=−kQqa2F_{12x} = -k\frac{Qq}{a^2}
- Rightward component from q4q_4: F14x=21kQ222a2F_{14x} = \frac{21kQ^2}{2\sqrt{2}a^2}
−kQqa2+21kQ222a2=0⇒−Qq+21Q222=0⇒q=21Q22-k\frac{Qq}{a^2} + \frac{21kQ^2}{2\sqrt{2}a^2} = 0 \Rightarrow -Qq + \frac{21Q^2}{2\sqrt{2}} = 0 \Rightarrow q = \frac{21Q}{2\sqrt{2}}
Y-direction:
- Downward force from q3q_3: F13y=−kQqa2F_{13y} = -k\frac{Qq}{a^2}
- Upward component from q4q_4: F14y=21kQ222a2F_{14y} = \frac{21kQ^2}{2\sqrt{2}a^2}
Same equation as x-direction, confirming consistency.
Final Answer:
qQ=2122≈7.42(unitless)\frac{q}{Q} = \frac{21}{2\sqrt{2}} \approx \boxed{7.42} \quad \text{(unitless)}
Explanation (300 words):
This problem involves balancing vector forces from Coulomb’s Law to determine when the net force on a charge is zero. Each particle interacts with every other, and because the configuration is a square, symmetry helps simplify the math. Charge q1=+Qq_1 = +Q experiences repulsive forces from other positive charges and attractive forces from the negative one. The aim is to set the vector sum of the electrostatic forces acting on q1q_1 to zero, considering both x and y components.
The force from each adjacent charge q2q_2 and q3q_3 is directly horizontal or vertical due to the square arrangement. Their forces are straightforward: equal in magnitude (since they are same charge qq and at equal distance aa), directed along the axes. However, the force from q4=−21Qq_4 = -21Q is along the diagonal and must be broken into x and y components using trigonometry (cos(45°) or sin(45°) = 1/√2).
By equating the sum of x-components and y-components of the forces to zero, we derive two identical equations involving qq and QQ. Solving for q/Qq/Q gives the ratio that ensures balance. This means that the magnitude of force from the very strong negative charge q4q_4 balances the combined repulsive forces of the smaller positive charges q2q_2 and q3q_3.
Hence, the ratio qQ=2122≈7.42\frac{q}{Q} = \frac{21}{2\sqrt{2}} \approx 7.42 ensures that the vector sum of all electrostatic forces on charge 1 is zero, meaning it is in equilibrium.