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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?

Sridev Venkataraman , 7 Years ago
Grade 12th Pass
anser 1 Answers
Samyak Jain
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 = \int_{(x1,y1,z1)}^{(x2,y2,z2)}  E.dr  ,  where dr = dx i + dy j + dz k
\therefore Potential difference between (–1,2) and (0,0) is V(–1,2) – V(0,0) = \int_{(0,0)}^{(-1,2)}  (5i + 15j).(dx i + dy j)
V(–1,2) – V(0,0) = \int_{(0,0)}^{(-1,2)}  (5dx + 15dy)   =  \[5x + 15y\]_{(0,0)}^{(-1,2)}  = – 5 + 30 = 25 V
i.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) = \int_{(0,0)}^{(2,-1)}  (5dx + 15dy)   =  \[5x + 15y\]_{(0,0)}^{(2,-1)}  =  10 – 15  = – 5 V.
\therefore V(2,–1) – (– 50 V) = – 5 V  \Rightarrow  V(2,–1) = (–5 – 50) V
    V(2,–1) = – 55 V.
Last Activity: 7 Years ago
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