Electric Fields of Like Charges

In summary, two small metallic spheres with a mass of .20g each and a string length of 30.0 cm are given equal electric charges and reach equilibrium when suspended as pendulums at a 5 degree angle with the vertical. Using the equations F=k(q1q2)/(r^2) and F=ma, the charge on each sphere can be determined by solving for the electrostatic force and vacuum permitivity. Newton's second law states that the total force on each sphere is zero in equilibrium, with forces including tension in the string, Earth's gravitational attraction, and electrostatic repulsion. By setting up equations for each axis and solving for the electrostatic force, the charge on each sphere can be determined. Ultimately
  • #1
Apple123
20
0
Can anybody help me with this problem?

Two small metallic spheres, each with a mass of .20g, are suspended as pendulums by light strings from a common point. They are given the same electric charge, and the two come to equilibrium when each string is at an angle of 5 degrees with the vertical. If the string is 30.0 cm long, what is the magnitude of the charge on each sphere.
 
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  • #2
Well,what are your thoughts so far...?Post ideas,calculations.

Daniel.
 
  • #3
Well, first i used the equation F=k(q1q2)/(r^2), but I can't figure out the value that the F would be. I used F=ma and try to find it that way, but the numbers don't seem to work out when i plug them into the equation. For q1 and q2 i just combined those, and then at the end after i solved for them, i divided the charge into two to get half a charge, which is the charge for one? I am having trouble, our teacher really hasn't explained this stuff yet, but they give us the assignment.
 
  • #4
The system is in equilibrium.Agree ?That means that the 2 charges are stationary/they don't move,ergo the total/resulting force on each of them is zero.

Take one charge.What forces act on it...?

Daniel.
 
  • #5
I think its gravity and the the other charge. Right?
 
  • #6
Nope.There are 3.

1.Tension in the string. [itex] \vec{T} [/itex]
2.Earth's gravitational attraction force. [itex] \vec{G} [/itex]
3.Electrostatic repulsion force. [itex] \vec{F}_{el} [/itex].

So what does that equilibrium mean ...?(HINT;Think Newton's second postulate).

Daniel.
 
  • #7
Newton's second law says F=ma, so in equalibrium that means that the sphere has to exert an opposite force for every force it encounters. The thing is i have no clue how to solve for them. I think i can solve for the attraction to earth, but i have no idea how to solve for the tension or repulsion force. Any hints?
 
  • #8
Equilibrium for one of the 2 spheres means

[tex]m\vec{a}=\vec{T}+\vec{G}+\vec{F}_{el}=\vec{0} [/tex]

Choose a system of 2 orthormal axis of coordinates and project that vector eq. on them.

Daniel.
 
  • #9
I don't understand what you are saying. This problem is really confusing me.
 
  • #10
I found this equation online, does it work?
http://homepage.smc.edu/physsci/dept/Physics/P22Hwk/EField.htm

Go there, its on number 5. Its the equation where q=2Lsin...

If so, how is it derived?
 
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  • #11
I'm trying to make you do it.I won't solve it for you,but help you do it yourself,as it is not my problem.

Let's pick Ox parallel to the ground pointing right and Oy perpendicular pointing upwards.

Ox:[tex] 0=-F_{el}+T\sin 5{}\mbox{deg} [/tex]

Oy:[tex] 0=-mg+T\cos 5{}\mbox{deg}[/tex]

Solve for F_{el}.Then u know that

[tex] F_{el}=\frac{q^{2}}{4\pi\epsilon_{0}\epsilon_{r}(2\cdot 0.3\sin 5{}\mbox{deg})^2} [/tex]

Daniel.
 
  • #12
Ok i understand every part to the problem except the 4Pi and the things after it up until the parentheses. What do those mean, and where did they come from.
 
  • #13
From Coulomb's law.Does it sound familiar...?(It should).

Daniel.
 
  • #14
You're probably familiar with

[tex] F_e = \frac{kq_1q_2}{r^2} [/tex]

What you might not know is 'k' is defined as

[tex] k = \frac{1}{4\pi\epsilon_0}[/tex]
 
  • #15
In the k equation, what is the symbol that follows the 4pi?
 
  • #17
ohhh ok i get it. Now to solve for F electric can u just use F=ma? And then for the two vacuum permitivities, are they the same number, or are they different?
 
  • #18
No.Just find it from the 2 eqns i posted and then use its expression to find the absolute value of the charge.

Daniel.
 
  • #19
Ohhh, so would u first solve Oy for T, then sub T into Ox and solve for F then take the absolute value? Then when u solve for the next equation, are the two vacuum permitivites both 8.85.10^12? I've never heard of this concept before.
 
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  • #20
After i did that equations, and the plugged them into the third equation, I got the value of 2.15x10^-14, does this seem about right or did i get off somewhere?
 
  • #21
The value for [itex] \epsilon_0 [/itex] is [itex] 8.85 \times 10^{-11} [/tex]

You could just use 9 x 10 ^ 9 for k, which is its value for vacuum. The error I believe is less than 1%.
 

1. What is an electric field?

An electric field is a physical quantity that describes the force exerted on a charged particle by other charged particles in its surroundings. It is represented by vectors and is measured in units of newtons per coulomb (N/C).

2. How are electric fields created by like charges?

Like charges repel each other, creating an electric field that pushes them away from each other. The strength of this field depends on the magnitude of the like charges and the distance between them.

3. What is the relationship between electric field strength and distance?

The electric field strength decreases as the distance from the like charges increases. This is because the force of repulsion between the like charges decreases with distance, resulting in a weaker electric field.

4. Can electric fields of like charges cancel each other out?

No, electric fields of like charges cannot cancel each other out. This is because they both exert a repulsive force on each other, resulting in a net force that is greater than zero.

5. How can we calculate the electric field of like charges?

The electric field of like charges can be calculated using Coulomb's law, which states that the electric field strength is directly proportional to the product of the like charges and inversely proportional to the square of the distance between them. It can also be calculated using the formula E = kQ/r^2, where k is the Coulomb's constant, Q is the charge of the like charges, and r is the distance between them.

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