A charge 10^{-9} C is located at the origin | vedclass

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(B) Let the charge at the origin be $q = 10^{-9} \ C$ and the charge at $(2, 0, 0) \ m$ be $Q$.Let point $P$ be $(3, 1, 1) \ m$.The position vector of

Vedclass · Accepted Answer

$-0.1424 	imes 10^{-9} \ C$

Vedclass · Answer

(B) Let the charge at the origin be $q = 10^{-9} \ C$ and the charge at $(2, 0, 0) \ m$ be $Q$.Let point $P$ be $(3, 1, 1) \ m$.The position vector of $P$ relative to the origin $O(0, 0, 0)$ is $\vec{r}_1 = (3 - 0)\hat{i} + (1 - 0)\hat{j} + (1 - 0)\hat{k} = 3\hat{i} + \hat{j} + \hat{k}$.The magnitude is $r_1 = |\vec{r}_1| = \sqrt{3^2 + 1^2 + 1^2} = \sqrt{11} \ m$.The position vector of $P$ relative to the charge $Q$ at $(2, 0, 0)$ is $\vec{r}_2 = (3 - 2)\hat{i} + (1 - 0)\hat{j} + (1 - 0)\hat{k} = \hat{i} + \hat{j} + \hat{k}$.The magnitude is $r_2 = |\vec{r}_2| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \ m$.The total electric field at $P$ is $\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{kq}{r_1^3}\vec{r}_1 + \frac{kQ}{r_2^3}\vec{r}_2$.The $Y$-component of the electric field is $E_y = \frac{kq}{r_1^3}(y_1) + \frac{kQ}{r_2^3}(y_2) = 0$.Substituting the values: $k \left[ \frac{10^{-9} 	imes 1}{(\sqrt{11})^3} + \frac{Q 	imes 1}{(\sqrt{3})^3} ight] = 0$.$\frac{Q}{3\sqrt{3}} = -\frac{10^{-9}}{11\sqrt{11}}$.$Q = -10^{-9} 	imes \frac{3\sqrt{3}}{11\sqrt{11}} = -10^{-9} 	imes \left( \frac{3}{11} ight)^{1.5} \approx -0.1424 	imes 10^{-9} \ C$.

$A$ charge $10^{-9} \ C$ is located at the origin of a coordinate system and another charge $Q$ at $(2, 0, 0) \ m$. If the $Y$-component of the electric field at $(3, 1, 1) \ m$ is zero,calculate the value of $Q$.

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