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Electromagnatism - Find charge deos this sphere have

  • Thread starter uniidiot
  • Start date
24
0
1. Homework Statement
The centre of a small sphere carrying a positive charge of 1.0 x 10^-8 C is fixed at 3cm above the centre of a small conducting ball, which had a negative charge and a mass of 0.05g. If the ball remains suspended, what charge does it have?

Answer is [ 4.9 x 10^-9]

but i can't seem to get that answer


2. Homework Equations



3. The Attempt at a Solution

Because the ball is suspended i take it that the force between the two charged bodies must be equal but opposite to the force due to gravity so:

F = mg

F = 0.05 x 9.8 = 0.49

F = (1/(4 x pi x epsilon 0)) x ((Q1 Q2) / r^2)

(1/(4 x pi x epsilon 0)) = 9 x 10^9

F = 9 x 10^9 x (Q1 Q2 / r^2)

Q2 = 9 x 10^9 x (Q1 / (r^2 x F)

Q2 = 9 x 10^9 x (10^-8 / (3^2 x 0.49)

Q2 = 20.4



please help :confused::confused::confused:
 

Doc Al

Mentor
44,803
1,061
F = mg

F = 0.05 x 9.8 = 0.49
Check your units.

F = (1/(4 x pi x epsilon 0)) x ((Q1 Q2) / r^2)

(1/(4 x pi x epsilon 0)) = 9 x 10^9

F = 9 x 10^9 x (Q1 Q2 / r^2)
OK.

Q2 = 9 x 10^9 x (Q1 / (r^2 x F)
Not sure what you did here.

Solve the problem algebraically first. Set your two equations for F equal to each other and solve for Q. The last step is to plug in numbers.
 
24
0
Quote:
Originally Posted by uniidiot View Post
F = mg

F = 0.05 x 9.8 = 0.49
Check your units.
i see what i've done, it should be in Kg not g

so

F = 0.00005 x 9.8 = 4.91 x 10^-4


Quote:
F = (1/(4 x pi x epsilon 0)) x ((Q1 Q2) / r^2)

(1/) = 9 x 10^9

F = 9 x 10^9 x (Q1 Q2 / r^2)
OK.

Quote:
Q2 = 9 x 10^9 x (Q1 / (r^2 x F)
Not sure what you did here.

Solve the problem algebraically first. Set your two equations for F equal to each other and solve for Q. The last step is to plug in numbers.
ok well:

F = x ((Q1 Q2) / r^2)

so if i rearrange to get Q2

Q2 = (1/(4 x pi x epsilon 0)) x (Q1 / (r^2 x F)

yes???
 

Doc Al

Mentor
44,803
1,061
F = x ((Q1 Q2) / r^2)

so if i rearrange to get Q2

Q2 = (1/(4 x pi x epsilon 0)) x (Q1 / (r^2 x F)

yes???
No. Redo that algebra. How did you get Q2 on the left side? (Looks like you divided by Q2 on the right, but multiplied by Q2 on the left. You have to do the same operation to both sides of an equation.)
 
24
0
ah, stupid me!

so

Q2 = (1/(4 x pi x epsilon 0)) x ((Q1 x f) / r^2)

Q2 = (9 x 10^-9) x (((1.0 x 10^-8) x (4.9 x 10^-4)) / 3^2)

Q2 = 4.9 x 10^-3

i, getting closer but, the answer should be 4.9 x 10^-9

i thought it might be the units but they seem ok now
 

Doc Al

Mentor
44,803
1,061
algebra error

so

Q2 = (1/(4 x pi x epsilon 0)) x ((Q1 x f) / r^2)
This is not true. You are taking an equation like this:
F = A*B

And transforming it into this:
A = B/F (NO GOOD!)

(Imagine A & B are your Q1 & Q2)

But you should get:
1/A = B/F

Which can be inverted to be:
A = F/B

Try this. Redo your algebra as I suggested earlier.
 
24
0
ok, ignoring (1/(4 x pi x epsilon 0)) for the minute

F = ( Q1 Q2 )/ r^2

F / Q2 = Q1 / r^2

1 / Q2 = Q1 / (F x r^2)

Q2 = (F x r^2) / Q1
 

Doc Al

Mentor
44,803
1,061
You are back on track now.
 
90
0
just plug in the values with appropriate units now
 
24
0
Thanks for everyones help =)

but i still cant get the corect answer:

Q2 = ((F x r^2) / Q1) x (1/(4 x pi x epsilon 0))

Q2 = ((4.91x10^-4 x 3^2) / 10^-8) x (9x10^9) = 3.97x10^15

Above the question it is stated that
(1/(4 x pi x epsilon 0)) = 9x10^9

i still cant see how im not getting [ 4.9 x 10^-9 C ]
 

Doc Al

Mentor
44,803
1,061
Thanks for everyones help =)

but i still cant get the corect answer:

Q2 = ((F x r^2) / Q1) x (1/(4 x pi x epsilon 0))
Your "(1/(4 x pi x epsilon 0))" is in the wrong place. It should be right next to Q1.

If you let k = (1/(4 x pi x epsilon 0)), then you have:
mg = k Q1 Q2/r^2

Solve for Q2 again (so you see where the k goes) and then plug in the numbers.
 
90
0
try to do it on paper if you are not already
because writing things out clearly with appropriate symbols is much simpler
than saying (1/4 x pi x epsilon)) x (((F x r^2) / Q1))...........and so on
 

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