Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 249510 • Letter: C
Question
Coulomb's law for the magnitude of the force Fbetween two particles with chargesQ and Q?separated by a distance d is
|F|=K|QQ?|d2,
where K=14??0, and ?0=8.854×10?12C2/(N?m2) is the permittivity of free space.
Part A
What is the net force exerted by these two charges on a third charge q3 = 52.5 nC placed between q1and q2 at x3 = -1.080 m ?
Your answer may be positive or negative, depending on the direction of the force.
Consider two point charges located on the x axis: one charge, q1 = -12.0 nC , is located atx1 = -1.735 m ; the second charge,q2 = 34.5 nC , is at the origin (x=0.0000).
Forces in a Three-Charge System Part A Part A What is the net force exerted by these two charges on a third charge g-52.5 nC placed between Coulomb's law for the magnitude of the force F between two particles with charges and separated by a distance d is and at =-1.080 ? Your answer may be positive or negative, depending on the direction of the force Express your answer numerically in newtons to three significant fiqures where K = , and 40 E0-8854 × 10 12 C2/(N-m2) is the permittivity of free space Force on q3 Consider two point charges located on the x axis one charge, q1 =-12.0 nC , is located at T1 1.135 m, the second charge q2-34.5 nC , is at the origin (z-0.0000) Submit Hints My Answers Give Up Review Part ide Feedbac ContinueExplanation / Answer
Since Q1 is negative and to the left of Q3 the force will be to the left
Since Q2 is positive and to the right of Q3 the the force will also be to the left
Therefore you can add the forces together (with left being negative)
Now we have F = -k*Q1*Q3/r13^2 - k*Q2*Q3/r23^2
= -9.0x10^9*12x10^-9*52.5x10^-9/(1.735 - 1.080)^2 - 9.0x10^9*34.5x10^-9*52.5x10^-9/1.080^2
= -7.59x10^-5N