Find Vab in an electric field E→=(2i^+3j^+4k^) N/C where r→a=(i^−2j^+k^)m and r→b=(2i^+j^−2k^)m

E→=2i^+3j^+4k^⇒V=-E→·dr→⇒V=−(2x+3y+4z)⇒Va=−(2−6+4)=0 and Vb=−(4+3−8)=+1⇒Vab=Va−Vb=0−1=−1 volt

Find Vab in an electric field E→=(2i^+3j^+4k^) N/C where r→a=(i^−2j^+k^)m and r→b=(2i^+j^−2k^)m

E→=2i^+3j^+4k^⇒V=-E→·dr→⇒V=−(2x+3y+4z)⇒Va=−(2−6+4)=0 and Vb=−(4+3−8)=+1⇒Vab=Va−Vb=0−1=−1 volt

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Find V_abin an electric field $\vec{E} = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) N / C$ where ${\vec{r}}_{a} = (\hat{i} - 2 \hat{j} + \hat{k}) m$ and ${\vec{r}}_{b} = (2 \hat{i} + \hat{j} - 2 \hat{k}) m$

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–1 volt

3 volt

2 volt

1 volt

answer is B.

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Detailed Solution

$\begin{array}{l} \vec{E} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \Rightarrow V = - \vec{E} \cdot d \vec{r} \Rightarrow V = - (2 x + 3 y + 4 z) \\ \Rightarrow V_{a} = - (2 - 6 + 4) = 0 \\ and V_{b} = - (4 + 3 - 8) = + 1 \\ \Rightarrow V_{ab} = (V_{a} - V_{b}) = 0 - 1 = - 1 volt \end{array}$