Two point dipoles p hat{k} and frac{p}{2} hat | vedclass

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(B) Let the dipole at origin be $\vec{p}_1 = p \hat{k}$ and the dipole at $(1, 0, 2)$ be $\vec{p}_2 = \frac{p}{2} \hat{k}$.We need to find the electri

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$\frac{-7p}{32 \pi \epsilon_0} \hat{k}$

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(B) Let the dipole at origin be $\vec{p}_1 = p \hat{k}$ and the dipole at $(1, 0, 2)$ be $\vec{p}_2 = \frac{p}{2} \hat{k}$.We need to find the electric field at $A(1, 0, 0)$.For dipole $\vec{p}_1$ at origin,point $A(1, 0, 0)$ lies on its equatorial line (since $\vec{p}_1$ is along $\hat{k}$ and the position vector of $A$ is $\hat{i}$). The electric field due to $\vec{p}_1$ is $\vec{E}_1 = -\frac{1}{4 \pi \epsilon_0} \frac{\vec{p}_1}{r_1^3} = -\frac{1}{4 \pi \epsilon_0} \frac{p \hat{k}}{1^3} = -\frac{p}{4 \pi \epsilon_0} \hat{k}$.For dipole $\vec{p}_2$ at $(1, 0, 2)$,the position vector of $A(1, 0, 0)$ relative to the dipole is $\vec{r}_2 = (1-1)\hat{i} + (0-0)\hat{j} + (0-2)\hat{k} = -2\hat{k}$. The distance is $r_2 = 2 	ext{ m}$. Since $\vec{p}_2$ is along $\hat{k}$ and the point $A$ lies on the axial line of $\vec{p}_2$,the electric field is $\vec{E}_2 = \frac{1}{4 \pi \epsilon_0} \frac{2 \vec{p}_2}{r_2^3} = \frac{1}{4 \pi \epsilon_0} \frac{2 (p/2) \hat{k}}{2^3} = \frac{p}{4 \pi \epsilon_0} \frac{\hat{k}}{8} = \frac{p}{32 \pi \epsilon_0} \hat{k}$.The resultant field is $\vec{E} = \vec{E}_1 + \vec{E}_2 = -\frac{p}{4 \pi \epsilon_0} \hat{k} + \frac{p}{32 \pi \epsilon_0} \hat{k} = \frac{-8p + p}{32 \pi \epsilon_0} \hat{k} = -\frac{7p}{32 \pi \epsilon_0} \hat{k}$.

Two point dipoles $p \hat{k}$ and $\frac{p}{2} \hat{k}$ are located at $(0, 0, 0)$ and $(1 \text{ m}, 0, 2 \text{ m})$ respectively. The resultant electric field due to the two dipoles at the point $(1 \text{ m}, 0, 0)$ is

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