Coulomb\'s law for the magnitude of the force F between two particles with charg

Question

Coulomb's law for the magnitude of the force F between two particles with charges Q and Q' separated by a distance d is |F| = K|QQ'|/d^2, where K = 1/4 pi epsilon_0, and epsilon_0 = 8.854 Times 10^-12 C^2 /(N m^2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q_1 = -17.5 nC, is located at x_1 = -1.740 m; the second charge, q_2 = 36.5 nC, is at the origin (x = 0.0000). What is the net force exerted by these two charges on a third charge q_3 = 49.5 nC placed between q_1 and q_2 at x_3 = -1.120 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Explanation / Answer

Here ,

q1 = 017.5 nC at x1 = -1.74 m

q2 = 36.5 nC at x = 0 m

let force at x3 = -1.120 m

net force on q3

Fnet = - F1 - F2

for the magnitude of force q3

Fnet = F1 + F2

Fnet = 9*10^9 * 49.5 *10^-9 * 10^-9 *(17.5/(1.74 - 1.12)^2 + 36.5/(1.12)^2)

Fnet = 3.32 *10^-5 N

the net force is 3.32 *10^-5 N

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---