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        An electric field E = (5i + 15j) N / C exists in a region. If the potential at (-1m,2m) is half of that in origin, potential at (2m, -1m) in volt is?
6 months ago

Samyak Jain
333 Points
							E = (5i + 15j) N / C, V(–1,2) = (1/2) V(0,0)           ...(1)We know that potential difference between two points A(x1,y1,z1) and B(x2,y2,z2) i.e.VAB = VB – VA = $\dpi{80} \int_{(x1,y1,z1)}^{(x2,y2,z2)}$  E.dr  ,  where dr = dx i + dy j + dz k$\dpi{80} \therefore$ Potential difference between (–1,2) and (0,0) is V(–1,2) – V(0,0) = $\dpi{80} \int_{(0,0)}^{(-1,2)}$  (5i + 15j).(dx i + dy j)V(–1,2) – V(0,0) = $\dpi{80} \int_{(0,0)}^{(-1,2)}$  (5dx + 15dy)   =  $\dpi{80} $$5x + 15y$\dpi{80}$_{(0,0)}^{(-1,2)}$  = – 5 + 30 = 25 Vi.e. (1/2)V(0,0) – V(0,0) = 25  or  V(0,0)  =  – 50 V           [From (1)]Again to find potential at (2,–1), which is V(2,–1) ,V(2,–1) – V(0,0) = $\dpi{80} \int_{(0,0)}^{(2,-1)}$  (5dx + 15dy)   =  $\dpi{80} $$5x + 15y$\dpi{80}$_{(0,0)}^{(2,-1)}$  =  10 – 15  = – 5 V.$\dpi{80} \therefore$ V(2,–1) – (– 50 V) = – 5 V  $\dpi{80} \Rightarrow$  V(2,–1) = (–5 – 50) V    V(2,–1) = – 55 V.

6 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions