An electric field is expressed as $\vec{E}=2 \hat{i}+3 \hat{j}$. Find the potential difference $\left(V_{A}-V_{B}\right)$ between two points $A$ and $B$ whose position vectors are given by $r_{A}=\hat{i}+2 \hat{j}$ and $r_{B}=2 \hat{i}+\hat{j}+3 \hat{k}$. Mark modulus of potential…

Question

An electric field is expressed as $\vec{E}=2 \hat{i}+3 \hat{j}$. Find the potential difference $\left(V_{A}-V_{B}\right)$ between two points $A$ and $B$ whose position vectors are given by $r_{A}=\hat{i}+2 \hat{j}$ and $r_{B}=2 \hat{i}+\hat{j}+3 \hat{k}$. Mark modulus of potential difference.

Aicharya · Accepted Answer

$\begin{aligned} v_{f}-v_{i} &= -\int_{i}^{f}\left(E_{x} \hat{i}+E_{y} \hat{j}+E_{z} \hat{k}\right) \cdot(d x \hat{i}+d y \hat{j}+d z \hat{k})\ V_{B}-V_{A} &=-\left[\int_{i}^{f} E_{x} d x+\int_{i}^{f} E_{y} d y+\int_{i}^{f} E_{z} d z\right] \ V_{B}-V_{A} &=-\left[\int_{1}^{2} 2 d x+\int_{2}^{1} 3 d y\right] \ V_{B}-V_{A} &=-[2(2-1)+3(1-2)] \ V_{B}-V_{A} &=-[2-3]=1 V \end{aligned}$ Hence, $V_{A}-V_{B}=-1 V .$ Hence, ans is 1

An electric field is expressed as \(\vec{E}=2 \hat{i}+3 \hat{j}\). Find the potential difference…

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