Question

A positive charge 50μC is located in xy plane at a positive vector $\vec{r_{o} }$ = $4\hat{i} + 4\hat{j}$. What is the electric field strength $\vec{E}$ at a point whose position vector is $\vec{r}$ = $10\hat{i} - 4\hat{j}$. ($\vec{r_{o} }$ and $\vec{r}$ are expressed in meter)

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Answer 1

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Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

A positive charge $\sf{ 50 \mu C }$ is located in xy plane at a positive vector $\bf{ \overrightarrow{r_{o}}=4 \hat{i}+4 \hat{j} }$ . What is the electric field strength $\vec{E}$ at a point whose position vector is $\vec{r}=10 \hat{i}-4 \hat{j}$ .($\vec{r}_{o}$ and $\vec{r}$ are expressed in meter)

♣ ᴀɴꜱᴡᴇʀ :

$\bf{|\overrightarrow{O P}|=|\vec{r}|=\sqrt{(10-4)^{2}+(-4-4)^{2}}=\sqrt{36+64}=10 \text { units }}$

$\bf{\vec{r}=\vec{r}_{2}-\vec{r}_{1}=10 \hat{i}-4 \hat{j}-4 \hat{i}-4 \hat{j}=6 \hat{i}-8 \hat{j}}$

Electric field at P in vector form,

$\bf{\vec{E}_{p}=\dfrac{Q}{4 \pi \epsilon_{0}|\vec{r}|^{3}} \vec{r}=\dfrac{\left(50 \times 10^{-6}\right) \times\left(9 \times 10^{9}\right)}{10^{3}}(6 \hat{i}-8 \hat{j})}$

$\bf{=2700 \hat{i}-3600 \hat{j} \mathrm{N} / \mathrm{C}=2.7 \hat{i}-3.6 \hat{j} \mathrm{kN} / \bf{C}}$

Magnitude of electric field at P,

$\bf{\left|\vec{E}_{p}\right|=\dfrac{Q}{4 \pi \epsilon_{0}|\vec{r}|^{2}}=\dfrac{\left(50 \times 10^{-6}\right) \times\left(9 \times 10^{9}\right)}{10^{2}}}$

$\bf{\left|\vec{E}_{p}\right|=4.5 \times 10^{3} \mathrm{N} / \mathrm{C}}$