Electric Potential and Electric Energy

[totallyclone]

Homework Statement

Consider point a which is 65 cm noth of a -2.8μC point charge, and point b which is 80 cm west of the charge. Determine (a) V_ba=V_b-V_a, and (b) E_b-E_a (magnitude and direction)

Homework Equations

k=9x10^-9
V=kQ/r
V=kQ(1/r_b-1/r_a)
E=kQ/r²
E=kQ(1/r_b²-1/r_a²)

The Attempt at a Solution

(a) V_ba=V_b-V_a
=kQ(1/r_b-1/r_a)
=(9x10^-9)(-2.8x10^-6)(1/0.8-1/0.65)
=7.269x10^-15 V

(b) E_b-E_a
=kQ(1/r_b²-1/r_a²)
=(9x10^-9)(-2.8x10^-6)(1/0.8²-1/0.65²)
=2.027x10^-14 N/C

I may have mistakes. I wonder if it's right. So I found the magnitude for (b) but I'm not quite sure about the direction (since it has a negative magnitude)? :shy:

 

[Doc Al]

(b) E_b-E_a
=kQ(1/r_b²-1/r_a²)
=(9x10^-9)(-2.8x10^-6)(1/0.8²-1/0.65²)
=2.027x10^-14 N/C

I may have mistakes. I wonder if it's right. So I found the magnitude for (b) but I'm not quite sure about the direction (since it has a negative magnitude)? :shy:

Realize that E_b and E_a are vectors. You must subtract them as vectors.

 

[totallyclone]

So I have to separate them to their own X and Y components??

 

[Doc Al]

So I have to separate them to their own X and Y components??

Sure, that's one way of doing it.

 

[Nickie]

Your calculations for (a) and (b) are correct. For the direction of (b), since the magnitude is positive, it means that the electric energy at point b is greater than the electric energy at point a. This indicates that there is a decrease in electric potential energy as you move from point a to point b, meaning that the electric field is pointing towards the negative charge at point a. Therefore, the direction of the electric field at point b is towards the negative charge at point a.