Two point charges q_1 = sqrt{10} , mu C and q | vedclass

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(A) The position of $q_1$ is $\vec{r}_1 = (1, 0)$ and $q_2$ is $\vec{r}_2 = (4, 0)$. The point $P$ is at $(0, 3)$.The displacement vectors are $\vec{r

Vedclass · Accepted Answer

$(63\hat{i} - 27\hat{j}) 	imes 10^2$

Vedclass · Answer

(A) The position of $q_1$ is $\vec{r}_1 = (1, 0)$ and $q_2$ is $\vec{r}_2 = (4, 0)$. The point $P$ is at $(0, 3)$.The displacement vectors are $\vec{r}_{P1} = (0-1)\hat{i} + (3-0)\hat{j} = -\hat{i} + 3\hat{j}$ and $\vec{r}_{P2} = (0-4)\hat{i} + (3-0)\hat{j} = -4\hat{i} + 3\hat{j}$.The distances are $r_1 = \sqrt{(-1)^2 + 3^2} = \sqrt{10} \, m$ and $r_2 = \sqrt{(-4)^2 + 3^2} = 5 \, m$.The electric field is $\vec{E} = \frac{kq_1}{r_1^3}\vec{r}_{P1} + \frac{kq_2}{r_2^3}\vec{r}_{P2}$.Substituting the values: $\vec{E} = 9 	imes 10^9 	imes 10^{-6} \left[ \frac{\sqrt{10}}{(\sqrt{10})^3}(-\hat{i} + 3\hat{j}) + \frac{-25}{5^3}(-4\hat{i} + 3\hat{j}) ight]$.$\vec{E} = 9 	imes 10^3 \left[ \frac{1}{10}(-\hat{i} + 3\hat{j}) - \frac{1}{5}(-4\hat{i} + 3\hat{j}) ight]$.$\vec{E} = 9000 \left[ (-0.1 + 0.8)\hat{i} + (0.3 - 0.6)\hat{j} ight] = 9000 [0.7\hat{i} - 0.3\hat{j}] = (63\hat{i} - 27\hat{j}) 	imes 10^2 \, V/m$.

Two point charges $q_1 = \sqrt{10} \, \mu C$ and $q_2 = -25 \, \mu C$ are placed on the $x$-axis at $x = 1 \, m$ and $x = 4 \, m$ respectively. The electric field (in $V/m$) at a point $y = 3 \, m$ on the $y$-axis is,[ take $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \, Nm^2C^{-2}$ ]

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