An electric dipole of moment overrightarrow{p | vedclass

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(C) Given the dipole moment $\overrightarrow{p} = (-\hat{i} - 3\hat{j} + 2\hat{k}) 	imes 10^{-29} \; C \cdot m$ and position vector $\overrightarrow{

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$(\hat{i} + 3\hat{j} - 2\hat{k})$

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(C) Given the dipole moment $\overrightarrow{p} = (-\hat{i} - 3\hat{j} + 2\hat{k}) 	imes 10^{-29} \; C \cdot m$ and position vector $\overrightarrow{r} = \hat{i} + 3\hat{j} + 5\hat{k}$.Since $\overrightarrow{r} \cdot \overrightarrow{p} = (1)(-1) + (3)(-3) + (5)(2) = -1 - 9 + 10 = 0$,the point lies on the equatorial plane of the dipole.For a point on the equatorial plane,the electric field $\overrightarrow{E}$ is given by $\overrightarrow{E} = -\frac{1}{4\pi\epsilon_0} \frac{\overrightarrow{p}}{r^3}$.This implies that the electric field $\overrightarrow{E}$ is antiparallel to the dipole moment $\overrightarrow{p}$.Therefore,$\overrightarrow{E} \parallel -\overrightarrow{p}$.Since $\overrightarrow{p} = (-\hat{i} - 3\hat{j} + 2\hat{k}) 	imes 10^{-29}$,then $-\overrightarrow{p} = (\hat{i} + 3\hat{j} - 2\hat{k}) 	imes 10^{-29}$.Thus,the electric field is parallel to $(\hat{i} + 3\hat{j} - 2\hat{k})$.

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