20.0-Forces in a Three-Charge System Coulomb\'s law for the magnitude of the for

Question

20.0-Forces in a Three-Charge System

Coulomb's law for the magnitude of the force Fbetween two particles with charges Q and Qseparated by a distance d is

|F|=K|QQ|d2,

where K=140, and 0=8.854×1012C2/(Nm2) is the permittivity of free space.

Consider two point charges located on the x axis: one charge, q1 = -14.5 nC , is located atx1 = -1.750 m ; the second charge,q2 = 39.0 nC , is at the origin (x=0.0000).

Part A

What is the net force exerted by these two charges on a third charge q3 = 45.5 nCplaced between q1 and q2 at x3 = -1.225 m ?

Your answer may be positive or negative, depending on the direction of the force.

Explanation / Answer

given,

q1 = -14.5 * 10^-9 C

q2 = 39 * 10^-9 C

q3 = 45.5 * 10^-9 C

q1 is located in x1 = -1.75 m

q2 is located in x2 = 0 m

q3 is located in x3 = -1.225 m

force between two charfed particles = k * Q * Q' / d^2

let force between q1 and q3 be f1

f1 = 9 * 10^9 * (-14.5 * 10^-9) * 45.5 * 10^-9 / (1.75 - 1.225)^2

f1 = -2.15 * 10^-5 N

let force between q2 and q3 be f2

f1 = 9 * 10^9 * 39 * 10^-9 * 45.5 * 10^-9 / 1.225^2

f1 = 1.064 * 10^-5 N

net force = f1 + f2

net force = -2.15 * 10^-5 + 1.064 * 10^-5

net force = -1.086 * 10^-5 N

et force exerted by these two charges on a third charge = -1.086 * 10^-5 N

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