Electric potential:Potential difference of test charge

[yuminie]

Homework Statement

A positive test charge of 1.5uC is placed in an electric field, 10 cm from another charge of -5.0uC that is anchored in place. What is the potential difference between the initial position of the test charge, and a point 5 cm closer to the negative charge?

Relevant Equations

Ee=kq1q2/r
Δv=ΔEe/q
v=kq/r

Electric potential energy at initial:
Ee=kq1q2/r
=(9 ×10 ^9×1.5×10^-6×(-5)×10^-6)/0.1
=-0.675J
Electric potential energy at the closer point:
Ee=kq1q2/r
=(9 ×10^9×1.5×10^-6×(-5)×10^-6)/0.05
=-1.35J
Δv=ΔEe/q
=(-1.35+0.675)/1.5×10^-6
=4.5×10^5V

or:

Initial position:
v=kq/r
=9 ×10^9×1.5×10^-6/0.1
=13500v
closer position:
v=kq/r
=9 ×10^9×(-5)×10^-6/0.05
=-900000v
Δv=-900000-13500=-913500V

I am very confused on this question. Please send help.

 

[Doc Al]

What is the potential difference between the initial position of the test charge, and a point 5 cm closer to the negative charge?

Note that they are asking about the potential due to the field of the anchored charge, not potential energy.

 

[yuminie]

Note that they are asking about the potential due to the field of the anchored charge, not potential energy.

Hi, thank you for your reply, this is how I changed my answer:
Initial position:
v=kq/r
=9 ×10^9×(-5)×10^-6/0.1
=-450000v
closer position:
v=kq/r
=9 ×10^9×(-5)×10^-6/0.05
=-900000v
Δv=-900000+450000=-450000V
Since its due to the field of the anchored charge, the charge of the test charge would not be needed?
Thanks.

 

[Doc Al]

Since its due to the field of the anchored charge, the charge of the test charge would not be needed?

Yes, that's how I interpret the problem.

 

[yuminie]

Yes, that's how I interpret the problem.

Thank you so much!