Electrostatics 1st year engineering

In summary, the conversation discusses the ratio of Q/q needed for a rhombus with charges of Q and q at each corner to have an angle of 90, 60, or 45 degrees. The tension of the rods is also taken into account. The person analyzing the forces on one corner determines that the signs of the charges are important in determining the directions of the forces. The equation used to find the ratio of Q/q is (Q/q) * (Q/cos^2(alpha/2)) = (q/sin^2(alpha/2)) * cot(alpha/2) +/- 8Qcos(alpha/2).
  • #1
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at each corner of a rhobus , made of rods, each with a length of B, there are electric charges of "Q" and "q" so that each q charge is opposite a q charge and each Q charge opposite a Q charge. what does the ratio of Q/q need to be so that the angle (alpha) is

a)90 degrees
b)60 degrees
c)45 degrees
(take the tension of the rods into account)

i started up by taking one of the corners and analyzing the forces on it, since the shape is static, Ftotal=0.
i took the top Q charge and said, what are the forces acting on it,

the Coulumb force F(Q-Q) which pushes it upwards
the Coulumb force F(Q-q) which (depends on the charge)
the tension os the rods T

my problem is that i do not know which of the charges are negative(if any) or positive, so i cannot decide on the directions of the forces, the only one i do know is F(Q-Q) since Q must repell Q,

diagram below


diagram below
http://picasaweb.google.com/devanlevin/DropBox?authkey=Gv1sRgCL_4l4PpvP_YsQE#5311131135994735922 [Broken]
 
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  • #2
what i did so far is:

since i don't know the sign of q, the coulomb force can either repell or attract Q

F(QQ)=2Tcos(α/2) ± 2F(Qq)cos(α/2)

q will always repell q so T is the force which balance that
2Tsin(α/2) = F(qq)


F(QQ)=F(qq)cot(α/2) ± 2F(Qq)cos(α/2)

K[tex]\frac{QQ}{(2b*cos(alpha/2))^2}[/tex] = K[tex]\frac{qq}{(2b*sin(alpha/2))^2}[/tex]*cot(alpha/2) ± 2K[tex]\frac{Qq}{b^2}[/tex]cos(alpha/2)

(Q/q) * [tex]\frac{Q}{cos^2(alpha/2)}[/tex] = [tex]\frac{q}{sin^2(alpha/2)}[/tex]*cot(alpha/2) ± 8Qcos(alpha/2)

1st of all, is this alright?
2nd of all, how do i find Q/q from this?? how do i get the q's and Q's out of this equation?
 
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  • #3


Your approach is correct in analyzing the forces on one of the charges. However, you are correct in saying that you do not know the charges of Q and q, so you cannot determine the direction of the forces. In order to solve for the ratio of Q/q, you will need to consider the net torque on the rhombus.

First, let's consider the case where alpha is 90 degrees. In this case, the rhombus will be in equilibrium, meaning the net torque on the rhombus must be zero. This means that the sum of the torques on the rhombus must be equal to zero.

To find the torque on each charge, we can use the equation tau = r x F, where r is the distance from the point of rotation (in this case, the corner of the rhombus) to the point of application of the force.

For the top Q charge, the torque will be counterclockwise, since the Coulomb force F(Q-Q) is pushing it upwards and away from the point of rotation. The distance r for this torque will be B/2, since the charge is located at the midpoint of the rod.

For the bottom Q charge, the torque will be clockwise, since the Coulomb force F(Q-Q) is pushing it downwards and towards the point of rotation. The distance r for this torque will also be B/2.

For the top q charge, the torque will be clockwise, since the Coulomb force F(Q-q) is pushing it downwards and towards the point of rotation. The distance r for this torque will be B, since the charge is located at the end of the rod.

For the bottom q charge, the torque will be counterclockwise, since the Coulomb force F(Q-q) is pushing it upwards and away from the point of rotation. The distance r for this torque will also be B.

Now, we can set up the equation for the sum of the torques:

0 = (B/2)(F(Q-Q)) + (B/2)(F(Q-Q)) + B(F(Q-q)) + B(F(Q-q))

Since we know that F(Q-Q) = k(Q^2)/B^2 and F(Q-q) = k(Qq)/B^2 (where k is the Coulomb constant), we can substitute these values into our equation:

0 = (B/2)(
 

1. What is electrostatics?

Electrostatics is the study of stationary electric charges and their interactions. It deals with the behavior of objects that have a net electric charge and the electric fields that are created by these charges.

2. What are the basic principles of electrostatics?

The basic principles of electrostatics include the concept of electric charge, Coulomb's law, electric fields, electric potential, and capacitance. These principles help us understand the behavior of electric charges and their interactions.

3. How is electrostatics applied in engineering?

Electrostatics is applied in engineering in various ways, such as in the design and operation of electronic circuits, in the development of electrostatic discharge (ESD) protection measures, in the production of inkjet printers and photocopiers, and in the design of electrostatic precipitators for air pollution control.

4. What are some common applications of electrostatics?

Some common applications of electrostatics include electrostatic painting, electrostatic separation of materials, electrostatic motors, and electrostatic loudspeakers. It is also used in everyday devices such as computer monitors, photocopiers, and air purifiers.

5. How can electrostatics be harmful?

Electrostatics can be harmful in certain situations, such as in the case of electrostatic discharge (ESD) which can damage electronic components. It can also cause discomfort or even injury to individuals in the form of electric shocks. Additionally, electrostatic charges can cause fires or explosions in certain environments, such as in the presence of flammable gases or dust.

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