Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 583611 • Letter: C
Question
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 = -16.0 nC , is located atx1 = -1.725 m ; the second charge,q2 = 39.5 nC , is at the origin (x=0.0000).
What is the net force exerted by these two charges on a third charge q3 = 51.5 nC placed between q1and q2 at x3 = -1.210 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
The magnitude of the net force on charge q3 is
Fnet = F13 + F23
where
Hence
Fnet = F13 + F23
Fnet = ( K q1 q3 ) / x132 + ( K q2 q3 ) / x232
Fnet = K q3 [ q1 / x132 + q2 / x232 ]
Fnet = ( 9x109N.m2/C2 )( 51.5x10-9C )[ ( 16.0x10-9 C ) / ( 0.515 m )2 + ( 39.5x10-9 C ) / ( 1.210 m )2 ]
Fnet = 4.05x10-5 N
The charge q1 exerts a force of attraction and q2 a force of repulsion. Both forces are to the left then, the net force is to the left, too