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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.
Notice that if we have a system of two particle placed in a epsilon medium then we can change system to vaccum by separating distance to epsilon*r where r is actual distance.
Therefore, the result follows
I think you can solve this question by calculating the equivalent permittivity between the two charges. Consider the charges Q1 and Q2 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.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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Comments
those two are same charged?positive or negative?
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it doesn't matter... say both are positive
F=K(Qq\r^2)
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two charges in two different medium, say Q is in air and q is in water ?
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.
Notice that if we have a system of two particle placed in a epsilon medium then we can change system to vaccum by separating distance to epsilon*r where r is actual distance. Therefore, the result follows
I think you can solve this question by calculating the equivalent permittivity between the two charges. Consider the charges Q1 and Q2 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.
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How?
force of attraction can be calculated as F=Qq4*3.142E divided by r^2