Problem Solving Strategies

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Hema

05:22

If a conducting sheet in a uniform electric field E has one side -Q and one side +Q (induced by presence of a similar sheet nearby), how is induced field equal to Q/Ae0?

John Rennie

@Hema The field inside a conductor is zero, so the induced field inside the plate has to exactly balance out the external field.

Hema

05:55

@JohnRennie ohhh ok

John Rennie

Suppose the voltage caused by the charge on the two sides of the plate is $V$, then the field this creates is $V/d$ where $d$ is the thickness of the plate.

If we treat the two surfaces of the plate as a capacitor then the capacitance is given by the usual parallel plate formula $C = \epsilon_0 A/d$

So if we have a charge $Q$ on the surfaces we get a voltage $V = Q/C = Qd/(\epsilon_0 A)$.

And $E = V/d = Q/(\epsilon_0 A)$

harambe

06:26

@JohnRennie good morning

John Rennie

@harambe morning :-)

Hema

07:06

@JohnRennie I just saw your message actually, got it thanks

John Rennie

@Hema Cool :-)

2 hours later…

harambe

08:49

@JohnRennie are you free now

John Rennie

@harambe hi, yes I'm around for a bit

harambe

Okay.. I will back in 10-20 min

harambe

09:23

@JohnRennie imgur.com/a/kbEV9gv

I tried using FBD and work energy theorem but it is not working

sammy gerbil

@Abcd Sorry my internet connection has been down for 16 hours.

harambe

@sammygerbil good morning

sammy gerbil

@harambe good morning

Can you explain how you are trying to solve the problem (#17)?

harambe

09:43

@sammygerbil let me post my attempt

imgur.com/a/1aSDpsB

@sammygerbil I got the value of angular velocity using this but I am confused for the latter part

I feel like using this value of angular velocity in com fbd.... But that will arise the problem of the hinge force unknowns

The centre of mass will have both centripetal as well as tangentisl acceleration

sammy gerbil

@harambe Your approach is correct. But there seems to be at least one error, in the first line. The distance of the COM of the rod alone is $l$ below the pivot not $2l$.

harambe

10:01

@sammygerbil on recalculating, I am getting distance of com to be 4/3 l below the pivot

sammy gerbil

@harambe The increase in GPE should be $4mgl$ I think.

harambe

Yea

I am getting the same thing

I got my mistake now. Okay

I think I got a sign mistake in my expression too. I will just try it once again

sammy gerbil

Your suggestion of finding centripetal and tangential accelerations of the COM is a good one. Vertically, the resultant force on the rod equals mass times acceleration of the COM. The rotational acceleration can be found from torque on the rod in this position.

harambe

@sammygerbil got it. I can handle this up now

sammy gerbil

ok. I will be online for a while if you need further help.

harambe

10:37

@sammygerbili am having difficulty in calculating vertical component of hinge force. I have cslculated angular acceleration but don't know the direction of tangentisl acceleration

I know angular velocity is decreasing so angular acceleration should be opposite direction of angular velocity......... I got it

I didn't see that the direction is given in the picture

harambe

10:55

@sammygerbil I got the answer now... This was tough. So basically we take advantage of the fact that the centre of mass is doing pure rotation around the hinged point and use our fbd which will contain the externsl forces acting on the centre of mass

sammy gerbil

@harambe Yes that is correct.

Hema

If two infinitely long line charges of same linear charge density are placed at an angle theta with each other(forming an x-like shape) what is electric field intensity at a point P which is at a distance x from point of crossing along angle bisector of line charges?

The answer given in my book is 4Kl/x where l is linear charge density

sammy gerbil

11:26

@Hema Have you tried to solve the question? Do you get a different answer?

You can use Gauss' Law to calculate the electric field due to each line of charge. The total field at P is then the vector sum of the electric field due to each line.

Hema

@sammygerbil by Gauss's law field due to a line charge at a point at a distance r is 2kl/r, here r = x sin (theta/2). We take the components of the two fields along the angle bisector which gives 2*(2kl/x sin (theta/2))*cos (theta/2)

This doesn't give the required answer

Is there something I'm doing wrong?

sammy gerbil

11:42

I don't see anything wrong with your calculation.

Hema

@sammygerbil but it gives (4kl/x) cot theta/2

And I'm supposed to get just 4kl/x

The next question has the same setup but with one line charge having the charge density -l and the answer is (4kl/x) cot theta/2

But shouldn't the magnitudes of both the answers be the same?

sammy gerbil

@Hema ok looks like the mistake is with the geometrical part of the calculation ... just nipping off, back in a few minutes.

Hema

@sammygerbil ohhh ok

But shouldn't both answers have the same magnitude as it is just the direction of the electric field due to the second line charge which becomes the opposite?

sammy gerbil

11:58

@Hema No not necessarily. The sum and difference of two vectors do not often have the same magnitude.

Hema

@sammygerbil ohhh ok

sammy gerbil

@Hema I think your factor of cos (theta/2) should actually be sin (theta/2). Then the answer when both wires have the same charge is 4kl/x.

Hema

@sammygerbil got it! And in the second case it is cos theta/2 right?

sammy gerbil

I think the factor of cot (theta/2) in the first case must be wrong because if the angle between the wires is small then the resultant electric field should be small, whereas cot (theta/2) becomes infinite for small angles theta.

Hema

@sammygerbil in the first case the two sin theta/2 's will cancel out to give 4kl/x right?

sammy gerbil

12:09

Yes a factor of cot (theta/2) in the 2nd case looks right.

Hema

@sammygerbil ohhh ok thank you!

2 hours later…

AbhigyanC

14:16

How to do SHM questions when two points are fixed?

For example: a rod is hinged to a point on a wall and a point on the rod is attached to a string attached to the ceiling

Then it is displaced perpendicular to the plane of both the rod and the string (into or out of the page)

How??

I'm well aware that we are supposed to displaced it slightly and look for the restoring torque, but I have no clue how to do that...

Abcd

14:37

@AbhigyanC try energy method once

@AbhigyanC please post the picture of full question

@AbhigyanC i want to try it (later) that's why i want the full pic

sammy gerbil

15:08

@AbhigyanC The fixed points AB define the axis about which the bob C is oscillating. It oscillates like a simple pendulum of fixed length $L=BC$. You need also to find the component of gravity $g'$ in the direction BC. Then the period of small oscillations is $2\pi \sqrt{L/g'}$.

@AbhigyanC Correction : the distance $L$ is not in general the distance BC but the perpendicular distance of the bob C from the axis AB. However, in this case BC is perpendicular to AB, so $L$ is the distance BC in this problem.

You must also take the component of $g$ along the same direction as $L$.

Anonymous

I do not understand the question Abhigyan asked, but will the string length remain constant?

Anonymous

I cannot understand how it can be displaced perpendicular to the plane without change in length of the string.

sammy gerbil

@IceInkberry The bob is given an impulse perpendicular to the plane of the page. The string does not change length. Looking at the motion in the plane of the page we will see the bob move towards B.

Anonymous

15:24

@sammygerbil Ah, I get it. It is given an impulse perpendicular to the plane of the rod and string and not exactly displaced perpendicular to it.

Anonymous

Thanks for clarifying!

sammy gerbil

@AbhigyanC Another important point is that I am assuming the rod has no mass. If it did have mass we would have to take into account its moment of inertia of the compund pendulum formed by the string, rod and bob about axis AB. The torque acting on it about axis AB would have to be calculated.

For small angles of displacement $\theta$ this torque is $Mxg'\theta$ where $M$ is the total mass of rod and bob, $x$ is the distance of the COM from axis AB and $g'$ is again the component of $g$ perpendicular to axis AB in the plane of the page.

sammy gerbil

15:59

If the rod makes an angle $\alpha$ with the axis of rotation AB then effectively it has length $2L\sin\alpha$ perpendicular to AB so its moment of inertia about axis AB is $\frac13 M (2L\sin\alpha)^2$.

Abcd

16:25

@sammygerbil Please let me know when you are free

sammy gerbil

@Abcd free now

Abcd

@sammygerbil till when?

sammy gerbil

@Abcd Well, I could die any minute now :) So if you have a question ...

Probably I shall be logged in to this chatroom for the next 6 hours or more.

Abcd

I am very tired right now. Will ask after 15-20 minutes