CK-12-Physics - Intermediate

15.2. Coulomb’s Law http://www.ck12.org

The forceF 1 is an attractive force placed on the 20μCpoint charge by the− 80 μCpoint charge.

F 1 =kq^20 r 1 q 280 =

(

8. 99 × 109 N∗m

2

C^2

)

( 20 × 10 −^6 C)( 80 × 10 −^6 C)

( 0. 50 m)^2 =^57.^5 →^58 N

The forceF 2 is a repulsive force placed on the 20μCby the 60μCpoint charge.

F 2 =kq^20 r 2 q 260 =

(

8. 99 × 109 N∗m

2

C^2

)

( 20 × 10 −^6 C)( 60 × 10 −^6 C)

( 0. 40 m)^2 =^67.^4 →^67 N

The angleθmust be determined in order to find the components of the forceF 1.

Using the tangent function we have tanθ=^00 .. 4030 mm→θ=tan−^100 ..^3040 mm= 36. 9 ◦→ 37 ◦.

The components ofF 1 :

F 1 x=F 1 sinθ= 67 .4 sin 37. 9 ◦= 41. 4 → 41 N

F 1 y=F 1 cosθ= 67 .4 cos 37. 9 ◦= 53. 2 → 53 N

The net force on the 20μCpoint charge in thex−direction is∑Fx=F 1 x= 41 N.

The net force on the 20μCpoint charge in they−direction is∑Fy=F 2 −F 1 y= 67 N− 53 N= 14 N.

The net force on the 20μCpoint charge is the Pythagorean sum of the components

√

412 + 142 = 43. 3 → 43 N

The direction of the force is tanθ=∑∑FFyx=^1441 NN→θ=tan−^11441 = 18. 9 ◦→ 19 ◦.

The final result can be written as< 43 N, 19 ◦>. Does the graphical sum of the vectorsF 1 and F 2 in the FBD above
agree with the mathematical result? (The vectors are not quite to scale!)