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# if two charged bodies are placed in two different medium , then how can we find the force of attraction between them ?

Note by Bhaskar Sen
3 years, 8 months ago

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I think you can solve this question by calculating the equivalent permittivity between the two charges. Consider the charges and separated by two mediums in series having permittivities epsilon 1 and epsilon 2. The effective permittivity, from my knowledge, would be 2 times epsilon1 * epsilon2 divided by (epsilon1 + epsilon2). Just put this in the Force equation for 2 electric charges with epsilon substituted as the effective epsilon, you'll get the answer. · 3 years, 8 months ago

I think you can solve this question by calculating the equivalent permittivity between the two charges. Consider the charges $$Q_{1}$$ and $$Q_{2}$$ separated by two mediums in series having permittivities epsilon 1 and epsilon 2. The effective permittivity, from my knowledge, would be 2 times epsilon1 * epsilon2 divided by (epsilon1 + epsilon2). Just put this in the Force equation for 2 electric charges with epsilon substituted as the effective epsilon, you'll get the answer. · 3 years, 8 months ago

Just a comment that would be a bit somehow related to my thread here, what is your concept of force due to charges in the presence of dielectrics? As you say q is in water and Q is in air, say q has a distance of $$x_1$$ from the water-air boundary, Q has a distance of $$x_2$$ from the water-air boundary. Consider another case, where q and Q are separated by air, with a 'slab' of water of 'thickness' $$x_1$$ in the middle (I know it's impossible but just imagine). The distance from q to Q is $$x_1+x_2$$. Will the answer differ in these 2 cases? If yes, how? · 3 years, 8 months ago

The 'slab' should be perpendicular to the line between Q and q, and just consider a 2D scenario

After this, using the fact that the force between two charges in a medium with relative permittivity $$\varepsilon$$, the force will be $$\displaystyle\frac{1}{4\pi\varepsilon_0\varepsilon}\frac{qQ}{r^2}$$, one can derive the answer. · 3 years, 8 months ago

Well the answer is, in both cases the answer is the same, as the 'thickness' of the two mediums are still the same. If we look at the formula for the force between two charges in a medium with relative permittivity $$\varepsilon$$, that is $$\displaystyle\frac{1}{4\pi\varepsilon_0\varepsilon}\frac{qQ}{r^2}$$. The difference of this with the force between two charges in vacuum, that is $$\displaystyle\frac{1}{4\pi\varepsilon_0}\frac{qQ}{r^2}$$, is the permittivity. So what is permittivity? Well loosely speaking it is the resistance that is encountered by an electric field. By this we can think this as a higher permittivity results a 'longer distance' between the charges. Here I approach the 'distance' $$(r)$$ as the space separating the two charges, and also the permittivity of the medium. So let's say the $$\text{distance}^2$$ in the case of vacuum only is $$r^2$$, and in the case of in a medium that has permittivity of $$\varepsilon$$, the $$\text{distance}^2$$ is $$r^2\varepsilon=(r\sqrt{\varepsilon})^2$$, which $$r$$ is dilated by a factor of $$\sqrt{\varepsilon}$$. So using a similar approach, the 'distance' in the case you stated will be $$x_1\sqrt{\varepsilon_1}+x_2\sqrt{\varepsilon_2}$$, where $$\varepsilon_1, \varepsilon_2$$ represent the relative permittivity of water and air. Substitute this as $$r$$ into the formula $$\displaystyle\frac{1}{4\pi\varepsilon_0}\frac{qQ}{r^2}$$, we get the answer $$\displaystyle\frac{1}{4\pi\varepsilon_0}\frac{qQ}{(x_1\sqrt{\varepsilon_1}+x_2\sqrt{\varepsilon_2})^2}$$. This is what I think though. · 3 years, 8 months ago

F=K(Qq\r^2) · 3 years, 8 months ago

two charges in two different medium, say Q is in air and q is in water ? · 3 years, 8 months ago

those two are same charged?positive or negative? · 3 years, 8 months ago

it doesn't matter... say both are positive · 3 years, 8 months ago