Question

Particles q1, 92, and q3 are in a straight line.

Particles q1 = -5.00 x 10-6 C,q2 = +2.50 x 10-6 C,
and q3= -2.50 x 10-6 C. Particles q1 and q2 are
separated by 0.500 m. Particles q and q3 are
separated by 0.250 m. What is the net force on q2?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
-5.00 x 10-6 C
91
0.500 m
+2.50 x 10-6 C
+92
-2.50 x 10-6 C
93
0.250 m

Answer 1

-1.35 N

Step-by-step explanation:

Force due to q1

• F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.5²
• F = 450 × 10⁻³
• F = 0.45 N (+)

Force due to q2

• F = 9 × 10⁹ × 5 × 10⁻⁶ × 2.5 × 10⁻⁶ / 0.25²
• F = 1800 × 10⁻³
• F = 1.8 N (-)

Net Force

• 0.45 - 1.8
• -1.35 N