A positive point charge 50 μC is located in the plane xy at a point with radius vector r→0=2i^+3j^. The electric field vector E→ at a point with radius vector r→=8i^−5j^,where r0 and rexpressed in meter, is

( 2.7 i ^ + 3.6 j ^ ) k N C − 1

A positive point charge 50 μC is located in the plane xy at a point with radius vector r→0=2i^+3j^. The electric field vector E→ at a point with radius vector r→=8i^−5j^,where r0 and rexpressed in meter, is

( 2.7 i ^ + 3.6 j ^ ) k N C − 1

( 1.4 i ^ + 2.6 j ^ ) k N C − 1

( 2.7 i ^ − 3.6 j ^ ) k N C − 1

( 2.7 i ^ + 3.6 j ^ ) k N C − 1

( 1.4 i ^ − 2.6 j ^ ) k N C − 1

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A positive point charge $50 μ C$ is located in the plane xy at a point with radius vector ${\vec{r}}_{0} = 2 \hat{i} + 3 \hat{j} .$ The electric field vector $\vec{E}$ at a point with radius vector $\vec{r} = 8 \hat{i} - 5 \hat{j},$ where $r_{0} a n d r$ expressed in meter, is

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$(1.4 \hat{i} + 2.6 \hat{j}) k N C^{- 1}$

$(2.7 \hat{i} - 3.6 \hat{j}) k N C^{- 1}$

$(2.7 \hat{i} + 3.6 \hat{j}) k N C^{- 1}$

$(1.4 \hat{i} - 2.6 \hat{j}) k N C^{- 1}$

answer is C.

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Detailed Solution

As $\vec{E} = \frac{1}{4 π ε_{0}} \frac{q}{{| \vec{r} - {\vec{r}}_{0} |}^{3}} (\vec{r} - {\vec{r}}_{0})$

$\vec{M N} = \vec{r} - {\vec{r}}_{0} = (8 \hat{i} - 5 \hat{j}) - (2 \hat{i} + 3 \hat{j}) = (6 \hat{i} - 8 \hat{j})$

$| \vec{r} - {\vec{r}}_{0} | = \sqrt{6^{2} + 8^{2}} = 10 m$

$\vec{E} = 9 \times 10^{9} \times \frac{50 \times 10^{- 6}}{{(10)}^{3}} (6 \hat{i} - 8 \hat{j}) E = (2.7 \hat{i} - 3.6 \hat{j}) k N C^{- 1}$

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