Coulomb's Law Exercises. Physics Homework Help, Examples.

Solved Physics Homework Examples, Problems on Coulomb's Law.

Our physics homework helper will email you your physics homework as shown with the solved examples shown below. Our helper also attends by fax upon request.

Exercise One, Homework Help Solutions. Coulomb's Law Applied to a Case of Electrical Interaction Between Equal Sign Charges.

Two small similar spheres with the same negative charge and 40 g of mass each are hanging at rest as indicated in the figure. The length of each string is 20 cm and the angle ß is 6º. Calculate how many electrons are needed to add to each sphere to obtain this situation.

The spheres distance is 2a and a is obtained from sin6º = a/0.2m, this is a = 0.021 m. The distance r between the spheres is then 0.042 m.

On the left sphere's free body diagram, where T is the string tension, F_s is the repulsed force the right sphere exerts over the left one and mg is the sphere weight. Then we have:

1) Sum of F_x = Tsinß - Fs = 0
2) Sum of F_y = Tcosß - mg = 0 or T = mg/cosß, which replaced in 1) we get: F_s = (mg/cosß)sinß = mgtgß = (40x10^-3 kg)(9.8 m/s²)tan6º = 4.12x10^-2 N.

The charge q on each sphere is calculated using Coulomb's law,

F_s = k_e q² /r².

Replacing r for 0.042 m and F_s for 4.12x10^-2 N and solving for q², q² = F_s r²/k_e = (4.12x10^-2 N)( 0.042 m)²/9x10⁹ Nm²/C². Then q = 8.98x10^-8 C.

The number of electrons needed to add to each sphere is 8.98x10^-8 C/1.602x10^-19C = 5.6x10¹¹ electrons.

Exercise Two, Homework Help Solutions. Coulomb's Law Applied to a Case of Three Point Charges

Assume three point charges located at the vertex of a right triangle as shown in the figure, where q₁ = -80µC, q₂ = 50µC and q₃ = 70µC, distance AC = 30 cm, distance AB = 40 cm. Calculate the force over the charge q₃ due to the charges q₁ and q₂.

The force directions coincide with the lines joining each pair of point charges. The force that q₁ exerts over q₃, F₃₁ is of attraction. The force that q₂ exerts over q₃, F₃₂ is repulsive. This is shown in the figure where the forces F₃₁ and F₃₂ are vectorially added to obtain F₃. The distance between q₃ and q₁ is obtained from

(CB)² = (AC)² + (AB)² = (0.3 m)² + (0.4 m)^²

from where CB = 0.5 m

The magnitudes of such forces are:

F₃₁ = [(9x10⁹ Nm² /C²)(80x10^-6 C)(70x10^-6 C)]/(0.5 m)² = 201.6 N

F₃₂ = [(9x109 Nm²/C²)(50x10^-6 C)(70x10^-6 C)]/ (0.3 m)²= 350 N

It is convenient to use coordinates axis as indicated, placing the origin on the charge where we need to calculate the resultant force, in this case on q₃.

The resultant force is F₃ = F₃₁ + F₃₂ . Then in terms of components x and y:
F_3x = F_31x + F_32x.
F_3y = F_31y + F_32y.

F_31x = F₃₁cosß = (201.6 N)(0.4/0.5) = 161.3 N ; F_31y = - F₃₁senß =-(201.6 N)(0.3/0.5) = -121 N
F₃₂ = 0 ; F_32y = F₃₂ = 350 N.
F_3x= 161.3 N + 0 = 161.3 N ; F_3y = -121 N + 350 N = 229 N.

The magnitude of the resultant force F₃ is obtained from (F₃)² = (F_3x)² + (F_3y)², giving F₃ = 280 N. The angle this force makes with the x axis is calculated using tgµ = F_3y/ F_3x= 229/161.3 = 1.42, then µ = arctan1.42 = 54.8º.

Note: You can access detailed information about our physics homework help by clicking below on "Physics, Detailed Homework Scope".