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There are multiple ways to interpret Coulomb's law in quantum electrodynamics (QED). Interestingly, they don't lead to quite the same conclusion (but there is no inconsistency because they are not defined in the same way). The most commonly used way (that ACuriousMind refers to in his comment) consists in relating the notion of classical potential with that ...


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It is really a matter of combinations. Potential energy is a feature of a system, so between two particles there is one potential energy. The summation however, will cadd the potential energy between two particles twice (e.g., $q_1\phi(\mathbf{r}_2)$ and $q_2\phi(\mathbf{r}_1)$). Hence, the one half term has to be introduced so that the potential energy of ...


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The electrostatic force is indeed $$F = \frac{q_1\cdot q_2}{4\pi \epsilon_0 r^2}$$ Now we need to move the two charges closer to each other - so we are doing work against this force. Indeed, there is an initial force needed to accelerate the particle (start the motion); however, assuming that the particle comes to rest again when it gets to $d_2$, then ...


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We assume $q_2$ does not accelerate, so the net force on it is zero while it is moving. The "extra" forces needed to get it going and to stop it at the end cancel each other out upon integration.


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The diagram you need to draw is this: Now we can compute the forces $$F_1 = \frac{q\cdot q_2}{4\pi\epsilon_0 (a^2 + d^2)}$$ The horizontal component is given by $$F_{1h} = F_1 \cdot \frac{d}{h}$$ The vertical component is $$F_{1v} = F_1 \cdot \frac{a}{h}$$ and finally, the force between the two green charges is $$F_2 = \frac{q_1\cdot ...


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If you only found the leftward force on the particle, technically your answer is wrong (or at least obtained by invalid methods). The vertical component may have been small enough that it didn't really affect the magnitude, but it would appear it certainly affects the direction. So what you'll need to do it calculate the vertical component of the force as ...


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Its the force experience by two unit charges by each other when separated by a distance of 1 metre.


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Coulomb's constant has units of metres per Farad. It is not acceleration; it is oftern expressed in terms of the vacuum permitivity for a free vacuum $k_\text{e} = \frac{1}{4\pi\varepsilon_0}$.


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Your first two equations are effectively the same once you consider that $F = q E$. You should use the second one if you're concerned about the electric field in general, but the first one if you want to know what actually happens to a particle (i.e., what force a particle feels). When in doubt, you can usually do alright by working out first the field, and ...


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The general equation for electric field due to dipole at a point is KP/R^3[(1+3cos^2(thetha)]^1/2. When (thetha=90 it is equitorial line and when (theta=0) it is axial line from the above equation we get the required results ie KP/R^3 and 2KP/R^3


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The difference probably is that the graph for the gravitational potential is the one for a spherical mass distribution (or a sphere with a certain mass if you wish) and the electric one is given for a point charge. You could also draw the gravitational potential for a point mass, then it would look equivalent to your electrical potential, or the other way ...


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Motion The fundamental relationship between Coloumb's law and Newton's Law of Gravitation is Motion. To explain let us first take Einstein's E=MC2. Now we know that anything times one is itself. Therefore when mass is 1, E=the speed of light squared. Therefore we can see that each piece of mass is the storage of the speed of light squared. Now in geometry ...



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