8:42 AM
Hi sir @JohnRennie
Hi :-)
Q:Three point charges q 2q and 8q are to be placed on a straight line of 9cm. Find the positions where the charges should be placed so that the P.E of this system is minimum. Im not really sure how to start this
You want the two biggest charges to be the farthest apart as they will repel each other most strongly and have the highest PE. Yes?
Does the PE of this means the sum of the absolute potential energy near the three charges?..
Yes
The sum of the PEs of all the pairs.
8:47 AM
Okay
kq₁q₂/r₁₂ + kq₁q₃/r₁₃ + kq₂q₃/r₂₃
Isnt this difference in PE between two charges
By this I mean kq1q2/r12 or kq1q3/r13 or kq2q3/r23
They are asking for the PE of the system i.e. the work required to assemble the system starting with all the charges at infinity.
Okay
Have you got it now? I need to go out in 5 mins ...
8:56 AM
Oh im doing it..but I think ill be able to do it.
Thankyou:)
I'll be back in about ³⁄₄ of an hour. You're welcome to ping me if you want to discuss it.
This is what I got.

2 hours later…
10:37 AM
hi :)
@Pizza Hi :-)
whats up?
What kind of questions can be asked in this channel?
Anything you want.
ah ok!
It was originally created to help students studying for the Indian JEE exam, but anyone can ask anything here.
I saw you had posted in The h Bar. That tends to be used for advanced discussions - university level and above.
10:43 AM
I'm doing physics 1 at university
but they are probably argument
of high school
because I'm at the beginning
Which country?
thanks for the clarification anyway!
@JohnRennie italy
I'm in the UK.
nice
10:46 AM
Do you have any advice to give me ?
@BinkyMcSquigglebottom Hi
It's been forty years since I was a first year physics student, so any advice I have is a long time out of date :-)
@JohnRennie Also me ,in London
I live in Chester.
I will also have to do physics 2, electromagnetism
@JohnRennie you know MrBean?
10:51 AM
@BinkyMcSquigglebottom The Rowan Atkinson character?
I remember watching it on TV a long time ago.
Was It funny ?
It's not really my kind of humour ...
My pfp Is him
11:11 AM
During volcanic explosions, massive fragments can be violently ejected from the crater
of rock: these projectiles are called volcanic bombs. The figure shows a vertical section of the
Mount Fuji, Japan.
a) At what initial speed should it be ejected
from the blowhole A to a bomb, at an elevation of 35°
on the horizontal plane, to fall at the foot of the volcano
in B, which is at a horizontal distance d=9.40
km and at a vertical distance h=3.30 km?
Let's ignore air friction for the moment.
A) 287.533 m/s
Right ?
g=9,81
11:34 AM
@BinkyMcSquigglebottom I get about 255 m/s
11:49 AM
I used the range formula with $y_0\neq0$
9400=x^2Sin[2*35°]/2*9,81(1+Sqrt[1+2*9,81*3300/x^2(sin(2*35°))^2)]
I put this on wolfram
@JohnRennie
12:05 PM
This Is the photo of the exercise
I can send It for you
Wait
Nice
Thanks
12:27 PM
@BinkyMcSquigglebottom We know that the bomb has to travel a distance 9400m to the right, so the time it takes is given by s = ut i.e.
t = 9400/vx
where vx = cv cosθ
And we know it has to travel a vertical distance of -3300m, where the minus sign means downwards, and this is given by s = ut + ¹⁄₂at² so:
-3300 = vy.t - ¹⁄₂gt²
where vy = v sinθ and t is the time from above: t = 9400/vx
Putting it together we get:
-3300 = v sinθ × 9400/(v cosθ) - ¹⁄₂g 9400²/(v²cos²θ)
I think I used another formula
$x_G = \frac{v^2_0\sin 2\theta}{2g} \left( 1 + \sqrt{1 + \frac{2g \ y_0}{v^2_0 \sin^2 \theta} \ }$
This
x_g=9400
That's no good ?
@JohnRennie
I need to go, sorry :-(
can look at it tomorrow morning if you're still having problems with it.
12:43 PM
Ok, np
Thanks anyway

4 hours later…
5:10 PM
$$x_G = \frac{v^2_0\sin 2\theta}{2g} \left( 1 + \sqrt{1 + \frac{2g \ y_0}{v^2_0 \sin^2 \theta} \ }\right)$$