Find (VA–VB) in an electric field E-=(2i+3j+4k)NC−1 where rA=(i−2j+k)m rB=(2i+j−2k)m

VAB=VA−VB=E⋅rB−rA= (2i + 3j + 4k).(i + 3j -3k)=+2 + 9 -12 = -1V

Find (VA–VB) in an electric field E-=(2i+3j+4k)NC−1 where rA=(i−2j+k)m rB=(2i+j−2k)m

VAB=VA−VB=E⋅rB−rA= (2i + 3j + 4k).(i + 3j -3k)=+2 + 9 -12 = -1V

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Find (V_A–V_B) in an electric field $\bar{E} = (2 i + 3 j + 4 k) {NC}^{- 1}$ where $r_{A} = (i - 2 j + k) m r_{B} = (2 i + j - 2 k) m$

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– 1V

– 4V

– 6V

– 2V

answer is B.

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Detailed Solution

$V_{AB} = V_{A} - V_{B} = E \cdot (r_{B} - r_{A})$
= (2i + 3j + 4k).(i + 3j -3k)=+2 + 9 -12 = -1V

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Find (VA–VB) in an electric field E-=(2i+3j+4k)NC−1 where rA=(i−2j+k)m rB=(2i+j−2k)m

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Find (V_A–V_B) in an electric field $\bar{E} = (2 i + 3 j + 4 k) {NC}^{- 1}$ where $r_{A} = (i - 2 j + k) m r_{B} = (2 i + j - 2 k) m$