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

Option: 1

0


Option: 2

4v


Option: 3

6.4v


Option: 4

2.8v


Answers (1)

best_answer

\vec{r}_{A B}=\vec{r}_A-\vec{r}_B=(3 \hat{i}-2 \hat{j}) \mathrm{cm} .

And 

Now,\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}

 \mathrm{\vec{E} \cdot\left(\vec{r}_A-\vec{r}_B\right) =V_A-V_B }

                                      \mathrm{ =\frac{4 \hat{\imath}+4 \hat{\jmath}}{\sqrt{2}} \cdot(3 \hat{\imath}-2 \hat{\jmath})}

                                      \mathrm{ =\frac{12-8}{\sqrt{2}} \Rightarrow 2 \sqrt{2}=2 \cdot 8 \mathrm{~V}}

 

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Sayak

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