A test charge of [tex]$2.00 \times 10^{-6} C$[/tex] is placed halfway between a charge of [tex]$8.75 \times 10^{-6} C$[/tex] and another of [tex]$2.50 \times 10^{-6} C$[/tex] separated by 10.0 cm. What is the magnitude of the force on the test charge?
Answer (1)
Calculate the force F 1 on the test charge due to q 1 using Coulomb's law: F 1 = 8.99 × 1 0 9 ( 0.05 ) 2 ∣ ( 8.75 × 1 0 − 6 ) ( 2.00 × 1 0 − 6 ) ∣ = 62.93 N .
Calculate the force F 2 on the test charge due to q 2 using Coulomb's law: F 2 = 8.99 × 1 0 9 ( 0.05 ) 2 ∣ ( 2.50 × 1 0 − 6 ) ( 2.00 × 1 0 − 6 ) ∣ = 17.98 N .
Calculate the net force on the test charge: F n e t = ∣ F 1 − F 2 ∣ = ∣62.93 − 17.98∣ = 44.95 N .
The magnitude of the net force on the test charge is 44.95 N .
Explanation
Problem Setup We are given three charges: a test charge q t = 2.00 × 1 0 − 6 C , and two other charges q 1 = 8.75 × 1 0 − 6 C and q 2 = 2.50 × 1 0 − 6 C . The charges q 1 and q 2 are separated by a distance of 10.0 cm, and the test charge q t is placed halfway between them. Our goal is to find the magnitude of the net force on the test charge.
Calculating Force F1 First, let's calculate the force F 1 on the test charge due to the charge q 1 . Coulomb's law states that the force between two point charges is given by F = k r 2 ∣ q 1 q 2 ∣ where k = 8.99 × 1 0 9 N m 2 / C 2 is Coulomb's constant, q 1 and q 2 are the magnitudes of the charges, and r is the distance between them. In our case, the distance between q t and q 1 is d 1 = 0.05 m . Therefore, F 1 = ( 8.99 × 1 0 9 N m 2 / C 2 ) ( 0.05 m ) 2 ∣ ( 8.75 × 1 0 − 6 C ) ( 2.00 × 1 0 − 6 C ) ∣ F 1 = ( 8.99 × 1 0 9 ) 0.0025 17.5 × 1 0 − 12 N = 62.93 N The force F 1 is repulsive since both charges are positive, and it acts along the line connecting q 1 and q t , directed away from q 1 .
Calculating Force F2 Next, let's calculate the force F 2 on the test charge due to the charge q 2 . The distance between q t and q 2 is d 2 = 0.05 m . Therefore, F 2 = ( 8.99 × 1 0 9 N m 2 / C 2 ) ( 0.05 m ) 2 ∣ ( 2.50 × 1 0 − 6 C ) ( 2.00 × 1 0 − 6 C ) ∣ F 2 = ( 8.99 × 1 0 9 ) 0.0025 5.00 × 1 0 − 12 N = 17.98 N The force F 2 is also repulsive since both charges are positive, and it acts along the line connecting q 2 and q t , directed away from q 2 .
Calculating Net Force Since q t is placed between q 1 and q 2 , the forces F 1 and F 2 act in opposite directions. The net force on q t is the difference between the magnitudes of F 1 and F 2 :
F n e t = ∣ F 1 − F 2 ∣ = ∣62.93 N − 17.98 N ∣ = 44.95 N The net force is directed away from the larger charge, q 1 .
Final Answer Therefore, the magnitude of the net force on the test charge is approximately 44.95 N.
Examples
Understanding the forces between electric charges is crucial in many real-world applications. For instance, in inkjet printers, charged droplets of ink are precisely controlled by electric fields to create images on paper. Similarly, electrostatic precipitators use electric forces to remove particulate matter from exhaust gases, reducing air pollution. The principles of Coulomb's law and net force calculations are also fundamental in designing and analyzing electronic devices, such as capacitors and transistors, which rely on the manipulation of electric fields and charges.
