319. Find the potential between two points p(1,-1,0) and q(2,1,3) with E = 40xy i + 20×2 j + 2 k
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c Explanation: V = -∫ E.dl = -∫ (40xy dx + 20x2 dy + 2 dz) , from q to p. On integrating, we get 106 volts.
c
See lessExplanation: V = -∫ E.dl = -∫ (40xy dx + 20×2 dy + 2 dz) , from q to p.
On integrating, we get 106 volts.
C Explanation: V = -∫ E.dl = -∫ (40xy dx + 20x2 dy + 2 dz) , from q to p. On integrating, we get 106 volts.
C
See lessExplanation: V = -∫ E.dl = -∫ (40xy dx + 20×2 dy + 2 dz) , from q to p.
On integrating, we get 106 volts.
Answer: c Explanation: V = -∫ E.dl = -∫ (40xy dx + 20x2 dy + 2 dz) , from q to p. On integrating, we get 106 volts.
Answer: c
See lessExplanation: V = -∫ E.dl = -∫ (40xy dx + 20×2 dy + 2 dz) , from q to p.
On integrating, we get 106 volts.
Answer: c Explanation: V = -∫ E.dl = -∫ (40xy dx + 20x2 dy + 2 dz) , from q to p. On integrating, we get 106 volts.
Answer: c
See lessExplanation: V = -∫ E.dl = -∫ (40xy dx + 20×2 dy + 2 dz) , from q to p.
On integrating, we get 106 volts.
c) Explanation: V = -∫ E.dl = -∫ (40xy dx + 20x2 dy + 2 dz) , from q to p. On integrating, we get 106 volts.
c)
See lessExplanation: V = -∫ E.dl = -∫ (40xy dx + 20×2 dy + 2 dz) , from q to p.
On integrating, we get 106 volts.