Coulomb's law hw question,

[emuhlay08]

q1= 2microC and q2=-4microC they are on a straight line 5.0 cm away from each other.

a)find the magnitude of the net force on charge q2 due to q1
b) in what direction will the net force acting on charge q2 due to q1 be directed
c)what is the magnitude and direction of the electric field at a point located 5 cm directly below charge q1
d)what is the magnitude of the electric potential located 5 cm directly below charge q1

I'm confused on what to do for only two points. Any help is appreciated and please explain so I can try to get an understanding

for A) i got 28.8 N and B) the forces will attract. not sure if those are correct

 

[STEMucator]

a) Looks fine.

b) Indeed the forces attract. On a free body diagram, which way would the force $\vec F_{12}$ point, given an arbitrary coordinate axis placed at $q_1$?

c) How do you calculate the electric field at a point, due to a point charge? Hint: Draw a free body diagram for this.

 

[emuhlay08]

a) Looks fine.

b) Indeed the forces attract. On a free body diagram, which way would the force $\vec F_{12}$ point, given an arbitrary coordinate axis placed at $q_1$?

c) How do you calculate the electric field at a point, due to a point charge? Hint: Draw a free body diagram for this.

This is what I came up with, it doesn't seem correct to me, but I'm not sure where I went wrong

 

[STEMucator]

This is what I came up with, it doesn't seem correct to me, but I'm not sure where I went wrong

You need to sum the components of the electric field of $\vec E_1$ and $\vec E_2$.

What about the direction? Draw $\vec E_{net_x}$ and $\vec E_{net_y}$ head to tail and apply $tan(\theta) = \frac{E_{net_y}}{E_{net_x}}$.

d) Same deal as c) pretty much, except potential is used instead of the electric field.

 

[HalfLostAndSearching]

.

For this problem, we can use Coulomb's law, which states that the magnitude of the force between two point charges is given by F = k(q1q2)/r^2, where k is the Coulomb's constant (9x10^9 Nm^2/C^2), q1 and q2 are the magnitudes of the two charges, and r is the distance between them.

a) To find the magnitude of the net force on charge q2 due to q1, we can simply plug in the values given into the Coulomb's law equation. So, we have F = (9x10^9 Nm^2/C^2)(2x10^-6 C)(-4x10^-6 C)/(0.05 m)^2 = -28.8 N. The negative sign indicates that the forces are attractive.

b) Since the net force is attractive, it will be directed towards q1, which is the direction of the negative charge.

c) To find the electric field at a point located 5 cm directly below charge q1, we can use the equation E = kq/r^2, where E is the electric field, k is the Coulomb's constant, q is the charge, and r is the distance between the point and the charge. In this case, we only have one charge, q1, so we can use the magnitude of this charge (2x10^-6 C). So, E = (9x10^9 Nm^2/C^2)(2x10^-6 C)/(0.05 m)^2 = 288 N/C. The direction of the electric field will be towards the negative charge, which is downwards in this case.

d) To find the electric potential at a point located 5 cm directly below charge q1, we can use the equation V = kq/r, where V is the electric potential, k is the Coulomb's constant, q is the charge, and r is the distance between the point and the charge. So, V = (9x10^9 Nm^2/C^2)(2x10^-6 C)/(0.05 m) = 360 V. The direction of the electric potential will be towards the negative charge, which is downwards in this case.