There exists two point charges of charges \(\text{q } C\) and \(\text{-q } C\). They are surrounded by an infinite conducting medium whose resistivity is \(\rho\). If they are placed \(d\) distance apart, find the time taken for the charges to become half their initial values. Report answer in terms of given parameters and known fundamental constants.

## Comments

Sort by:

TopNewestTry this – Karan Shekhawat · 1 year, 6 months ago

Log in to reply

Thanks for Posting..... I'am trying – Karan Shekhawat · 1 year, 6 months ago

Log in to reply

EDIT-

the current is \(\frac {q }{\rho \epsilon} \) and this equals \( \frac {-dq}{dt} \)

solving which yields time,

The method is to choose the symmetry plane of the charges,

and use E/p = j (current density), now proceed – Mvs Saketh · 1 year, 6 months ago

Log in to reply

– Raghav Vaidyanathan · 1 year, 6 months ago

Sorry, the question was to find the time taken for the charges to become half their initial value. I got mixed up. The method is right. \[J=\sigma E\]Log in to reply

(hint- its not complicated) – Mvs Saketh · 1 year, 6 months ago

Log in to reply

– Krishna Sharma · 1 year, 6 months ago

Hey! I remember that you have posted exactly this type of problem some time ago.Log in to reply

– Mvs Saketh · 1 year, 6 months ago

yes i did, and i deleted as i accidentally uploaded with wrong answer and was too lazy to correct itLog in to reply

\[B=\frac { \sigma { \mu }_{ 0 }q }{ 2\pi { \varepsilon }_{ 0 }r } \left[ 1-\frac { a }{ \sqrt { { a }^{ 2 }+{ r }^{ 2 } } } \right] \]

where \(a=d/2\) , \(r\) is distance of point from the axis on the plane of symmetry. – Raghav Vaidyanathan · 1 year, 6 months ago

Log in to reply

Luckily – Mvs Saketh · 1 year, 6 months ago

Log in to reply

– Raghav Vaidyanathan · 1 year, 6 months ago

I dunno man, I used ampere's law and ohm's law to get the above equation...Log in to reply

@Mvs Saketh

How did you calculate current? Did you integrate?? – Raghav Vaidyanathan · 1 year, 6 months ago

Log in to reply

– Mvs Saketh · 1 year, 6 months ago

yesLog in to reply

allfield lines that begin at + chargemustterminate at - charge?? I am doing Ronak's new problem right now.. It is very interesting.. I thought sun's rays are parallel when they hit the earth.. – Raghav Vaidyanathan · 1 year, 6 months agoLog in to reply

Well, For each field line that starts from positive charge and goes to infinity, there is a field line that does the reverse for the negative charge and carries absolutely the same current density at all points

So, we can simply connect them at infinity, to complete the circuit which is what i have done when integrating from 0 to infinity

Yes, its an amazing problem, and you are right, but remember that you can only approximate them as paralell when there is no first order deviation , that is why you will face trouble if you try to solve the uniform field case with a single charge,

Try using a more hoziontal situation(charge system) instead that has no vertical field uptill first order in any deviations (cause you are gonna integrate it later, and first order term will become finite)(i hope you got what i mean) THink over it, there are 3 methods so far i know to solve it, you will see the discussion in solutions

and yes, you can solve it using symmetry as well, and superposition, there is a popular iridov problem that is some what like that, the resistance grid one– Mvs Saketh · 1 year, 6 months agoLog in to reply

– Raghav Vaidyanathan · 1 year, 6 months ago

Method of reflection is too damn difficult... I will try this problem later. As for the current discussion, I want to know whether we can apply Gauss' law to find out the flux through the central plane of symmetry. The expression looks so simple that It makes me wonder if it is possible. I have a method, but I think it becomes inconsistent when applied to cases when the charges are unequal in magnitude.Log in to reply

@Raghav Vaidyanathan -

The flux is \(\frac {q}{\epsilon} \)

Consider one charge, half of the flux passes through the plane

so flux contribution \( \frac {q}{2\epsilon}\)

and similarly for the other,

adding em both you get the result – Mvs Saketh · 1 year, 6 months ago

Log in to reply

Are we considering them as separate and then using superposition?? If so will this also be valid for charges with magnitude \(2q,-q\)? – Raghav Vaidyanathan · 1 year, 6 months ago

Log in to reply

And yes, it will work for unequal charges as well, try it

But remember that if you are thinking of using flux to find current, then your answer wont match because now more of the current is escaping/coming from infinity, so upon the symmetry plane, there will be an additional component of current density that is vertically upward/downward since they no longer cancel out – Mvs Saketh · 1 year, 6 months ago

Log in to reply

– Raghav Vaidyanathan · 1 year, 6 months ago

Yes, this is exactly what I wanted to know. I understand now. Thanks a lot!Log in to reply

– Karan Shekhawat · 1 year, 6 months ago

Sorry how did you got expression of current ... what element you are taking ..? How you are proceeding ? Yes I believe J=(sigma)E will definately use , what path should I choose ...Log in to reply

use formula to find current density, then choose concentric rings, integrate over them till infinity to arrive at answer – Mvs Saketh · 1 year, 6 months ago

Log in to reply

@Karan Shekhawat @Mvs Saketh – Raghav Vaidyanathan · 1 year, 6 months ago

Log in to reply