Two charges of -4 mu C and +4 mu C are plac | vedclass

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(B) The dipole moment $\vec{p}$ is given by $\vec{p} = q(\vec{r}_B - \vec{r}_A)$.Given $q = 4 \ \mu C = 4 	imes 10^{-6} \ C$,$\vec{r}_A = (1, 0, 4) \

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$2$

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(B) The dipole moment $\vec{p}$ is given by $\vec{p} = q(\vec{r}_B - \vec{r}_A)$.Given $q = 4 \ \mu C = 4 	imes 10^{-6} \ C$,$\vec{r}_A = (1, 0, 4) \ m$,and $\vec{r}_B = (2, -1, 5) \ m$.$\vec{p} = 4 	imes 10^{-6} 	imes [(2-1)\hat{i} + (-1-0)\hat{j} + (5-4)\hat{k}] = 4 	imes 10^{-6} (\hat{i} - \hat{j} + \hat{k}) \ C \cdot m$.The electric field is $\vec{E} = 0.20 \ \hat{i} \ V/cm = 0.20 	imes 10^2 \ \hat{i} \ V/m = 20 \ \hat{i} \ V/m$.The torque $\vec{	au}$ is given by $\vec{	au} = \vec{p} 	imes \vec{E}$.$\vec{	au} = [4 	imes 10^{-6} (\hat{i} - \hat{j} + \hat{k})] 	imes [20 \hat{i}] = 80 	imes 10^{-6} [\hat{i} 	imes \hat{i} - \hat{j} 	imes \hat{i} + \hat{k} 	imes \hat{i}] \ Nm$.Since $\hat{i} 	imes \hat{i} = 0$,$\hat{j} 	imes \hat{i} = -\hat{k}$,and $\hat{k} 	imes \hat{i} = \hat{j}$,we have:$\vec{	au} = 80 	imes 10^{-6} [0 - (-\hat{k}) + \hat{j}] = 80 	imes 10^{-6} (\hat{j} + \hat{k}) \ Nm = 8 	imes 10^{-5} (\hat{j} + \hat{k}) \ Nm$.The magnitude of the torque is $|\vec{	au}| = 8 	imes 10^{-5} \sqrt{1^2 + 1^2} = 8 \sqrt{2} 	imes 10^{-5} \ Nm$.Comparing this with $8 \sqrt{\alpha} 	imes 10^{-5} \ Nm$,we get $\alpha = 2$.

Two charges of $-4 \ \mu C$ and $+4 \ \mu C$ are placed at the points $A(1, 0, 4) \ m$ and $B(2, -1, 5) \ m$ located in an electric field $\vec{E} = 0.20 \ \hat{i} \ V/cm$. The magnitude of the torque acting on the dipole is $8 \sqrt{\alpha} \times 10^{-5} \ Nm$. Where $\alpha = $ . . . . . .

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