Particles P1 and P2 are located in three dimensional space a

Particles P1 and P2 are located in three dimensional space at the points (2.00, 3.50, 1.75) mm and (3.50, 2.25, 2.00) mm. These particles carry charges of +3.00 C and +4.50 C respectively. What is the magnitude of the net force they exert on another particle P3, with charge 3.50 C, located at the origin?

Could you draw a diagram and include detailed formula\'s used to help me understand how to solve this type of question? Thank you!

Solution

Diagram is in 3-D, it will be confusing drawing.

We will deal this problem with cartesian vector notation.

P1 is +ve and P3 is -ve do force will be attractive.

force will be from P3 to P1

that means force F1 have direction along P3P1 vector.

P3P1 = P1 - P3

= (2.00i - 3.50j + 1.75k ) - (0i + 0j + 0k )

= 2.00i - 3.50j + 1.75k mm

magnitude = sqrt(2^2 + 3.50^2 + 1.75^2) = 4.395 mm Or 4.395 x 10^-3 m

electric force = (kq1q2/|r|^2 ) rcap

Or we can write , F = (kq1q2 / |r|^3) r vector

F1 = [(9 x 10^9 x 3 x 10^-6 x 3.50 x 10^-6) / (4.295 x 10^-3)^3 ] (2.00i -3.50j + 1.75k ) x 10^-3

F1 = 2385.46i - 4174.56j + 2087.28 k N   

Similarly :

P3P2 = P2 - P3

= (- 3.50i + 2.25j - 2.00k ) - (0i + 0j + 0k )

= - 3.50i + 2.25j - 2.00k mm

magnitude = sqrt(2^2 + 3.50^2 + 2.25^2) = 4.62 mm Or 4.62 x 10^-3 m

F = (kq1q2 / |r|^3) r vector

F1 = [(9 x 10^9 x 4.5 x 10^-6 x 3.50 x 10^-6) / (4.62 x 10^-3)^3 ] (- 3.50i + 2.25j - 2.00k ) x 10^-3

F1 = - 5042.43i +3241.56j - 2881.39 k N   

Fnet = F1 + F2

= - 2656.97i - 933j -794.11k

magnitude = sqrt(2656.97^2 + 933^2 + 794.11^2)

= 2925.85 N