Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1587801 • Letter: C
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|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 = -10.5 nC , is located at x1 = -1.735 m ; the second charge, q2 = 38.0 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 = 45.0 nC placed between q1 and q2 at x3 = -1.165 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
Electric field strength = Force / Charge (N/C)
Recall also the Superposition theorem which states that the E-field produced by a distribution of charges at a point, is equal to the sum of the E-fields from each individual charge. So
E = E1 + E2 + E3 + ........En
So what I'll do is forget about q3 for the moment and work out the strength of the E-field at q3's location due to the presence of q1 and q2. Once I have done that, I can just multiply that by the charge on q3 to work out the force acting on q3 at that location.
Let E1 = E-field produced by q1
Let E2 = E-field produced by q2
Then from Coulomb's law for the E-field [ remember this is not the FORCE law], we have
E1 = Kq1 / r1^2
E2 = Kq2/r2^2
r1 = 1.735 - 1.165 = 0.57 m
r2 = 1.165 m
E = [K * (- 10.5 * 10^-9) / 0.57^2] + [K * (38 * 10^-9) / 1.165^2]
= K [ (- 32.32 * 10^-9) + (28 * 10^-9)]
= K [ - 4.32 * 10^-9]
= 9 x 10^9* - 4.32 * 10^-9 = - 38.88 N/C
So the force on q3 is
= E * q3
= - 38.88 * 45* 10^-9
= 0.1750 * 10^-5 N
The minus sign just tells you that the force is acting in the direction of the negative x-axis which makes sense as q3 will be repelled by q2 [both positive] and attracted by q1.