Electric Fields of Like Charges

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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
dextercioby
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Well,what are your thoughts so far...?Post ideas,calculations.

Daniel.
 
  • #3
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Well, first i used the equation F=k(q1q2)/(r^2), but I cant figure out the value that the F would be. I used F=ma and try to find it that way, but the numbers dont 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
dextercioby
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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
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I think its gravity and the the other charge. Right?
 
  • #6
dextercioby
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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
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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
dextercioby
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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
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I dont understand what you are saying. This problem is really confusing me.
 
  • #10
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I found this equation online, does it work?
http://homepage.smc.edu/physsci/dept/Physics/P22Hwk/EField.htm [Broken]

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

If so, how is it derived?
 
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  • #11
dextercioby
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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
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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
dextercioby
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From Coulomb's law.Does it sound familiar...?(It should).

Daniel.
 
  • #14
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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
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In the k equation, what is the symbol that follows the 4pi?
 
  • #17
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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
dextercioby
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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
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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? Ive never heard of this concept before.
 
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  • #20
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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
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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%.
 

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