Electric Potential and Electric Energy

AI Thread Summary
The discussion focuses on calculating the electric potential difference (Vba) and the electric field difference (Eb-Ea) between two points near a -2.8μC point charge. The calculations yield Vba as approximately 7.269x10^-15 V and the magnitude of Eb-Ea as 2.027x10^-14 N/C. There is uncertainty regarding the direction of the electric field difference due to its negative magnitude. Participants emphasize the importance of treating electric fields as vectors and suggest breaking them into X and Y components for accurate direction determination. The conversation highlights the need for careful vector analysis in electric field calculations.
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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) Vba=Vb-Va, and (b) Eb-Ea (magnitude and direction)

Homework Equations


k=9x10-9
V=kQ/r
V=kQ(1/rb-1/ra)
E=kQ/r2
E=kQ(1/rb2-1/ra2)

The Attempt at a Solution


(a) Vba=Vb-Va
=kQ(1/rb-1/ra)
=(9x10-9)(-2.8x10-6)(1/0.8-1/0.65)
=7.269x10-15 V

(b) Eb-Ea
=kQ(1/rb2-1/ra2)
=(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:
 
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totallyclone said:
(b) Eb-Ea
=kQ(1/rb2-1/ra2)
=(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 Eb and Ea are vectors. You must subtract them as vectors.
 
So I have to separate them to their own X and Y components??
 
totallyclone said:
So I have to separate them to their own X and Y components??
Sure, that's one way of doing it.
 
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