A point charge 50 μC is located in the XY plane at the point of position vector r→0=2i^+3j^. What is the electric field at the point of position vector r→=8i^-5j^

Q=50×10−6; r0→=2i+3j and r→=8i−5jr→−r0→=6i−8j=E→=14π∈0Qr→−r0→2⇒9×109×50×10−662+822⇒450×103100=4500V/m

A point charge 50 μC is located in the XY plane at the point of position vector r→0=2i^+3j^. What is the electric field at the point of position vector r→=8i^-5j^

Q=50×10−6; r0→=2i+3j and r→=8i−5jr→−r0→=6i−8j=E→=14π∈0Qr→−r0→2⇒9×109×50×10−662+822⇒450×103100=4500V/m

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A point charge 50 $μC$ is located in the XY plane at the point of position vector ${\vec{r}}_{0} = 2 \hat{i} + 3 \hat{j}$ . What is the electric field at the point of position vector $\vec{r} = 8 \hat{i} - 5 \hat{j}$

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900V/m

4500 V/m

0.04V/m

1200V/m

answer is D.

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Detailed Solution

$\begin{array}{l} Q = 50 \times 10^{- 6}; \vec{r_{0}} = 2 i + 3 j and \vec{r} = 8 i - 5 j \\ \vec{r} - \vec{r_{0}} = 6 i - 8 j = \vec{E} = \frac{1}{4 π \in_{0}} \frac{Q}{{|\vec{r} - \vec{r_{0}}|}^{2}} \\ \Rightarrow \frac{9 \times 10^{9} \times 50 \times 10^{- 6}}{{(\sqrt{6^{2} + 8^{2}})}^{2}} \\ \Rightarrow \frac{450 \times 10^{3}}{100} = 4500 V / m \end{array}$