# HCV2 pg 122 Q31

## HCV2 pg 122 Q31

Two particles A and B each having a charge Q are placed distance d apart . Where should a particle of charge q be placed on the perpendicular bisector of AB so that it experiences maximum force?

What is the magnitude of the force?

What is the magnitude of the force?

**Shohini**- Posts : 15

Join date : 2008-05-31

Age : 27

## Re: HCV2 pg 122 Q31

Let's put the charges on the x axis at locations x= -d/2 and x=+d/2. Then by symmetry, the x component of the E field will cancel out to zero. The total force will depend on the size of the y component of E. Since both charges are equal, this will just be double the size of E_y of one charge.

Therefore, the problem becomes: find the maximum value of E_y on the y axis. Since E = kQ/r^2, and since E_y/E = y/r by similar triangles, this means that E_y = kQyr^-3. The maximum will occur where d/dy(E_y) = 0.

Calculating the derivative gives d/dy(kQyr^-3) = kQ[r^-3 -3yr^-4*dr/dy]. Since this should be zero, it must be true that r^-3 = 3yr^-4*dr/dy. Now multiply through by r^4 to ger r = 3ydr/dy.

Next, what is dr/dy? The easy way is to compute dr^2/dy two different ways: dr^2/dy = 2rdr/dy, but also dr^2/dy = d((d/2)^2 + y^2)dy = 2y. Therefore, 2rdr/dy = 2y, so dr/dy = y/r. Therefore, you get r = 3ydr/dy = 3y^2/r, so r^2=3y^2. But r^2 = (d/2)^2 + y^2, so this means that (d/2)^2 + y^2 = 3y^2, so (d/2)^2 = 2y^2. Therefore, y = d/(2sqrt(2)).

Here is another way to do the calculation, using angles. One charge Q is at point A(-d/2,0), and the particle is at B(0,y). Let a be the angle BAO between the line AB and the x axis. Then r = d/2*sec(a), so E = kQ/(d/2*sec(a))^2 and E_y/E = sin(a). Therefore, E_y = kQsin(a)/(d/2*sec(a))^2 = 4kQ/d^2*(cos(a))^2*sin(a). This will be maximum when 0 = d(E_y)/da = 4kQ/d^2[-2cos(a)(sin(a))^2 + (cos(a))^3]. Since this is zero, it must be true that 2cos(a)(sin(a))^2 = (cos(a))^3. Dividing both sides by (cos(a))^3 gives 2(tan(a))^2 = 1, so tan(a) = 1/sqrt(2). But y = d/2*tan(a), so y = d/(2sqrt(2). This way is a little shorter, but you have to choose a less obvious variable (a instead of y) to make it work.

You could probably do this same problem without using as many "fancy" derivative tricks as I used, but the formulas would be messier. I usually prefer, instead of putting all the details in at the beginning, to put in a few details, work on it a little bit, put in more details, work some more, and so on, until I get to the answer. This keeps every formula small and neat as you go along.

Hope this helps!

Pinnaka

Ps:Please do invite u r frnds to the forum

Therefore, the problem becomes: find the maximum value of E_y on the y axis. Since E = kQ/r^2, and since E_y/E = y/r by similar triangles, this means that E_y = kQyr^-3. The maximum will occur where d/dy(E_y) = 0.

Calculating the derivative gives d/dy(kQyr^-3) = kQ[r^-3 -3yr^-4*dr/dy]. Since this should be zero, it must be true that r^-3 = 3yr^-4*dr/dy. Now multiply through by r^4 to ger r = 3ydr/dy.

Next, what is dr/dy? The easy way is to compute dr^2/dy two different ways: dr^2/dy = 2rdr/dy, but also dr^2/dy = d((d/2)^2 + y^2)dy = 2y. Therefore, 2rdr/dy = 2y, so dr/dy = y/r. Therefore, you get r = 3ydr/dy = 3y^2/r, so r^2=3y^2. But r^2 = (d/2)^2 + y^2, so this means that (d/2)^2 + y^2 = 3y^2, so (d/2)^2 = 2y^2. Therefore, y = d/(2sqrt(2)).

Here is another way to do the calculation, using angles. One charge Q is at point A(-d/2,0), and the particle is at B(0,y). Let a be the angle BAO between the line AB and the x axis. Then r = d/2*sec(a), so E = kQ/(d/2*sec(a))^2 and E_y/E = sin(a). Therefore, E_y = kQsin(a)/(d/2*sec(a))^2 = 4kQ/d^2*(cos(a))^2*sin(a). This will be maximum when 0 = d(E_y)/da = 4kQ/d^2[-2cos(a)(sin(a))^2 + (cos(a))^3]. Since this is zero, it must be true that 2cos(a)(sin(a))^2 = (cos(a))^3. Dividing both sides by (cos(a))^3 gives 2(tan(a))^2 = 1, so tan(a) = 1/sqrt(2). But y = d/2*tan(a), so y = d/(2sqrt(2). This way is a little shorter, but you have to choose a less obvious variable (a instead of y) to make it work.

You could probably do this same problem without using as many "fancy" derivative tricks as I used, but the formulas would be messier. I usually prefer, instead of putting all the details in at the beginning, to put in a few details, work on it a little bit, put in more details, work some more, and so on, until I get to the answer. This keeps every formula small and neat as you go along.

Hope this helps!

Pinnaka

Ps:Please do invite u r frnds to the forum

**Pinnaka**- Admin
- Posts : 36

Join date : 2008-05-20

Age : 31

Location : India

Page

**1**of**1****Permissions in this forum:**

**cannot**reply to topics in this forum