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...
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...