Recent content by yuminie

Thank you so much!

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...