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Torque on a Dipole

H4: H-4. F...

Question

H4:

H-4. Find out the magnitude of electric field intensity and electric potential due to a diplole of dipole moment $P = i + \sqrt{3 \bar{j}}$ $P = i + \sqrt{3 \bar{j}}$ kept at origin at following points.
(i) (2,0,0) (ii) (-1, $\sqrt{3}$ ,0)

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Solution

dear student electric field on a general point

$E = K P / r^{3} \sqrt{1 + 3 \cos^{2} θ} w h e r e θ = a n g l e b e t w e e n p o s i t i o n v e c t o r a n d d i p o l e m o m e n t v e c t o r V a t g e n e r a l p i n t V = K P \cos θ / r^{2} w h e r e θ = a n g l e b e t w e e n d i p o l e m o m e n t m o m e n t v e c t r a n d p o s i t i o n v e c t o r a t t h a t l o c a t i o n$

$I) p o t e n t i a l \cos θ = a n g l e b e t w e e n d i p o l e m o m e n t v e c t o r a n d o s i t i o n v e c t o r h e r e \vec{p} = i + \sqrt{3} j |\vec{p}| = 2 \vec{r} = 2 i |\vec{r}| = 2 \cos θ = \vec{p} . \vec{r} / |\vec{p}| |\vec{r}| \cos θ = \frac{i + \sqrt{3} j) . 2 i}{2 * 2} = 1 / 2 V = k p \cos θ / r^{2} = \frac{9 * 10^{9} * 2}{2 * 2 * 2} = 2.25 * 10^{9} V a g a i n w h e n \vec{r} = - i + \sqrt{3} j \cos θ = (- i + \sqrt{3} j) . (i + \sqrt{3} j) / (2 * 2) = 1 / 2 V = V = k p \cos θ / r^{2} = \frac{9 * 10^{9} * 2}{2 * 2 * 2} = 2.25 * 10^{9} V t o f i n d e l e c t r i c f i e l d a t \vec{r} = 2 i E = \frac{k p}{r^{3}} \sqrt{1 + 3 \cos^{2} θ} h e r e \cos θ = 1 / 2 E = \frac{9 * 10^{9} * 2}{2^{3}} \sqrt{1 + 3 * 1 / 4} = \frac{9 * \sqrt{7} * 10^{9}}{8} V / m e l e c t r i c f i e l d a t \vec{r} = - i + \sqrt{3} h e r e \cos θ = 1 / 2 s o E = \frac{9 * 10^{9} * 2}{2^{3}} \sqrt{1 + 3 * 1 / 4} = \frac{9 * \sqrt{7} * 10^{9}}{8} V / m$

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