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-1

You know what i want to say to you is that maths is not only about equation and number it is more about logic, so whenever we go from one step to other step we use logic and each step in any derivation tells something. What mathematicians have done is that they have given symbols to some phrases or word(like, = is symbolized to "is equals to"). so what i ...


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E=k*p/z^3. This is because usually, the Electric field is calculated at a point whose distance from the midpoint of the dipole(z) is very large as compared to distance between the two opposite charges(say a). That is z>>a. Thus, the original value of dipole at a point on the equatorial line=kp/(z^2+a^2)^3/2. as z^2+a^2=z^2.(z>>a). Thus, E=kp/z^3.


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A dipole may be represented by a pair of nearby charges of the opposite sign (but the same absolute values), $+Q$ and $-Q$ at a distance $\vec r$ from each other. At the distance $R$ from the dipole (and I will only consider the case when we measure the field on the axis that includes the dipole itself), one adds $E$ going like $+1/R^2$ from $+Q$, and ...


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A dipole has two parameters magnitude and orientation. If you change the orientation the nature of the force also changes (i.e from attractive to repulsive), so what you are saying is if you drop a body on suppose say North Pole and it is free falling towards earth then on the South Pole it should free fall away from earth which is not the case. ...


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Similar problem is two charged particles $q_1$ and $q_2$ separated by the distance d. The TOTAL energy is $E=q_1 \ f_2 = \frac{q_1 \ q_2}{d}$, where $f_2$ is potential of the second particle. Thus, you see that you take the charge of the first particle and multiply by the potential of the second one. It gives you the total energy of system.


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Q1. But how this can be explained "theoretically"? I assume your question is about the concept of 'electric potential ' due to a distribution of charges and in the present case 'a dipole'. The best way is to imagine an unit positive charge being carried/moved from infinity to a point on the equatorial line of the dipole. Naturally your probe charge ...


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If you look at the electric filed lines you will note that a positive charge moving in the direction of the blue arrow does have a force on it but that force is always at right angles to its motion. Hence no work is done moving the charge.



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