Question

You are working for a manufacturing company. Your supervisor has an idea for controlling the position of a small bead by using electric fields. The physical setup is shown in the figure below.

Answer 1

You are working for a manufacturing company, which is mathematically given as

• $m=3\sqrt{2}$
• $m=\frac{15\sqrt{5}}{16}$
• x=0.747a
• $m/n=\frac{(x^2+a^2)3/2}{x^3}$

What is the value of m that will place the movable bead in equilibrium at x-a a ....?

a)

Generally, the equation for the force of equilibrium is mathematically given as

F=2fcos\theta

Therefore

$K(npq^2/a^2)=2\frac{kmpq^2}{a\sqrt{2}^2}0.5\\\\np=mP/ \sqrt{2}\\\\where n=3$

$m=3\sqrt{2}$

b)

By force equilibrium

$K(npq^2/(2a^2))=2*\frac{kmpq^2}{a\sqrt{5a}^2}* \frac{2a}{\sart{5a}}$

Therefore

$n/4=2/5*m*2/\sqrt{5}\\\\m= \frac{5\sqrt{5}}{16}$

$m=\frac{15\sqrt{5}}{16}$

c)

$K(npq^2/x^2)=2\frac{kmpq^2}{a\sqrt{x^2+a^2}^2}0.5*x/\sqrt{x^2+a^2}$

x^2+a^2=(14/3)^{2/3}x^2

x=a/1.338

x=0.747a

d)

By force equilibrium

$K(npq^2/x^2)=2\frac{kmpq^2n}{\sqrt{x^2+a^{3/2}}^2}\\\\n/x^2=\frac{2mx}{(x^2+a^2)^{3/2}}$

$m/n=\frac{(x^2+a^2)3/2}{x^3}$

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