- #51

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ohh thanks buddy!@Ayesha02 if you haven't solved differential equations before, then suppose that ##f (t)=A\cos (\omega t-B)## and use the initial conditions I hinted at in a previous post to find the unknowns.

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- #51

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ohh thanks buddy!@Ayesha02 if you haven't solved differential equations before, then suppose that ##f (t)=A\cos (\omega t-B)## and use the initial conditions I hinted at in a previous post to find the unknowns.

- #52

- #53

etotheipi

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I agree, it's quite a fun question. Am interested also!@Ayesha02 can you please tell me from which book is this exercise , for me at least it represents a fine blending of electrostatics with circuit theory ,and I just want to know from which book it is.

- #54

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We have a real tough exam here that we've to appear, and this is one of the questions from our preparatory material.

- #55

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Probably from Irodov's problems in general physics or a JEE problems textbook.@Ayesha02 can you please tell me from which book is this exercise , for me at least it represents a fine blending of electrostatics with circuit theory ,and I just want to know from which book it is.

- #56

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You have written ##\epsilon_0 ## a wrong way.

I further think you copied wrong because that answer is dimensionally incorrect.

How about you left out a "1/"?

- #57

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yeah i got that...i missed the reciprocal

I further think you copied wrong because that answer is dimensionally incorrect.

How about you left out a "1/"?

Thatanswer would be correct.

although, could you help me understand the solution theyve given?

- #58

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The charge oscillates back and forth between the two spheres at that radian frequency.yeah i got that...i missed the reciprocal

although, could you help me understand the solution theyve given?

The only way you're going to get the answer is by solving the differential equation. There are a number of ways to do that, for example the way a previous poster suggesting a solution of the form ## cos(\omega t - \phi) ##, ##\phi ## a constant.

- #59

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Ohh finally!The charge oscillates back and forth between the two spheres at that radian frequency.

The only way you're going to get the answer is by solving the differential equation. There are a number of ways to do that, for example the way a previous poster suggesting a solution of the form ## cos(\omega t - \phi) ##, ##\phi ## a constant.

understood:)

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What have you got up until now from solving the differential equation?although, could you help me understand the solution theyve given?

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- #62

etotheipi

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Yeah, there are a few different ways you could think about it. Either a closed loop to infinity and back, in which case you can treat the potential as the potential difference between the sphere and infinity: ##\frac{kQ_A}{r} - L\frac{di}{dt} - \frac{kQ_B}{r}##. Or you could think about it in the sense of two potentials ##\frac{kQ_A}{r}## and ##\frac{kQ_B}{r}## and a potential difference between them...isan interesting probem because it makes concrete the nexus between 'potential' of physics and 'voltage' of electrical engineering.

Can blur the lines a little but but usually the distinction is that to discuss potential we have a fixed reference, so it means being extra careful when discussing the potential

- #63

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The most recent one i have put up goes by the title- 'Hard Momentum Conservation'

See u there:)

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etotheipi

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@etotheipi I need u there my man:)

The most recent one i have put up goes by the title- 'Hard Momentum Conservation'

See u there:)

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Sorry, I had to:

We badly need some humor on this site. We all love physics but we need to take even it a bit more lightly now and then. Thank you! (Of course this song always reminds me of 'Dr. Strangelove' too.)

- #67

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We helpers love to help! Really! Forces us to review also, often; so in a sense you're a helper too! And all of us love physics all the time!

The most recent one i have put up goes by the title- 'Hard Momentum Conservation'

See u there:)

- #68

etotheipi

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Especially now that cabin fever has fully set in

We badly need some humor on this site. We all love physics but we need to take even it a bit more lightly now and then. Thank you! (Of course this song always reminds me of 'Dr. Strangelove' too.)

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