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A uniform electric field of 400 \mathrm{~V} / \mathrm{m} is directed at 45^{\circ} above \mathrm{x}- axes as shown in figure. The potential difference \mathrm{V_{A}-V_{B}}

Option: 1

0


Option: 2

\mathrm{4v}


Option: 3

\mathrm{6.4v}


Option: 4

\mathrm{2.8 v}


Answers (1)

best_answer

\mathrm{\begin{aligned} & \vec{\gamma}_{A B}=\vec{\gamma}_A-\vec{r}_B=(3 \hat{i}-2 \hat{j}) \mathrm{cm} \\ & \vec{E}=4 \cos 45^{\circ} \hat{\imath}+4 \sin 45 \hat{\jmath}=\frac{4 \hat{\imath}+4 \hat{\jmath}}{\sqrt{2}} \mathrm{v} / \mathrm{cm} \end{aligned}}
Now, \mathrm{\vec{E} \cdot\left(\vec{\gamma}_A-\vec{\gamma}_B\right)=V_A-V_B}
\mathrm{\begin{aligned} & =\frac{4 \hat{\imath}+4 \hat{\jmath}}{\sqrt{2}} \cdot(3 \hat{\imath}-2 \hat{j}) \\ & =\frac{12-s}{\sqrt{2}} \Rightarrow 2 \sqrt{2}=2.8 \mathrm{~V} \end{aligned}}

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Shailly goel

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