1 μC charge is shifted from A to B and it is found that work done by an external force is 40 μJ in doing so against electrostatic forces then, find the magnitude of potential difference VB - VA (in volts).
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
answer is 40.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
The potential difference between two points in electric fieldVB−VA=Wext(A→B)q=40×10−61×10−6=40VHence VA−VB=−40V