• Support PF! Buy your school textbooks, materials and every day products Here!

Finding charges using electric field and distance

  • Thread starter scholio
  • Start date
  • #1
160
0

Homework Statement


at a point midway between two point charges separated by 0.10 meters, the magnitude of the electric field is 10V/m and the electric potential is zero. what are the two charges


Homework Equations



electric potential - > point charge V = kq/r

electric potential difference = electric potential V = kq [(1/r_B) - (1/r_A)] where r is the distance, k is constant = 9*10^9, q is charge in coulombs

electric field E = qF where q is charge, F is force in newtons

force F = kq_1q_2/r^2


The Attempt at a Solution



r = 0.05m

I need to be able to isolate the two charges so that i can solve for one and then subsitute the value in to solve for the other.

if i use electric potential difference eq, how do i involve two charges? the radii should be the same, correct? but if radii are the same won't i get zero in the [] and then q = 0. also if i use that equation i will not be able to factor in electric field.

i am supposed to get q_A = -q_B = 1.39*10^-12 coulombs

help appreciated
 
Last edited:

Answers and Replies

  • #2
nrqed
Science Advisor
Homework Helper
Gold Member
3,568
189

Homework Statement


at a point midway between two point charges separated by 0.10 meters, the magnitude of the electric field is 10V/m and the electric potential is zero. what are the two charges


Homework Equations



electric potential - > point charge V = kq/r

electric potential difference = electric potential V = kq [(1/r_B) - (1/r_A)] where r is the distance, k is constant = 9*10^9, q is charge in coulombs

electric field E = qF where q is charge, F is force in newtons

force F = kq_1q_2/r^2


The Attempt at a Solution



r = 0.05m

I need to be able to isolate the two charges so that i can solve for one and then subsitute the value in to solve for the other.

if i use electric potential difference eq, how do i involve two charges? the radii should be the same, correct? but if radii are the same won't i get zero in the [] and then q = 0. also if i use that equation i will not be able to factor in electric field.

i am supposed to get q_A = -q_B = 1.39*10^-12 coulombs

help appreciated

I don't know why you wrote an expression for the potential difference. What you should impose is that the total potential (the sum) is zero. And yes, the two distances used in the two electric potential will be equal since the point is midway.

This will give a (very simple) relation between the two charges (but it won't give you the magnitude yet)

Next, write an expression for the total E field at the mid point.

(Note: the E field is the electric force divided by the charge, not qF as you wrote).

be careful with the electric field, it's a vector quantity so you must pay attention to the direction.
 
  • #3
160
0
thanks, you were right, finding the relationship between the two charges was exceedingly simple. i am having trouble finding the magnitude of the charge, here is what i have done:

relationship between charges q_A = -q_B

E = F/q

let F = k(q_B)(-q_B)/0.05^2 = (9*10^9)(-(q_B)^2)/(0.05^2) = -3.6*10^12 (q_B^2)
let E = 10, q = q_B

so i got 10 = -3.6*10^12 q_B^2 and q_B = -2.7 * 10 ^ -12 coulombs

i am supposed to get 1.39 * 10 ^ -12 coulombs

what did i do wrong?
 
Last edited:
  • #4
nrqed
Science Advisor
Homework Helper
Gold Member
3,568
189
thanks, you were right, finding the relationship between the two charges was exceedingly simple. i am having trouble finding the magnitude of the charge, here is what i have done:

relationship between charges q_A = -q_B

E = F/q

let F = k(q_B)(-q_B)/0.05^2 = (9*10^9)(-(q_B)^2)/(0.05^2) = -3.6*10^12 (q_B^2)
let E = 10, q = q_B

so i got 10 = -3.6*10^12 q_B^2 and q_B = -2.7 * 10 ^ -12 coulombs

i am supposed to get 1.39 * 10 ^ -12 coulombs

what did i do wrong?
?????????
The 10 V/m is the total E field. You seem to set only one E field equal to that value.

Don't introduce the force at all, directly use the E field (I don't see you dividing F by q before getting E!).

Find the E field produced by one charge, the E field produced by the second charge (assume that the positive charge is on the left and the negative on the right or vice versa), add the result, find the magnitude and set that equal to 10V/m.
Note: the expression for the magnitude of the E field produced by a point charge is simply [tex] E = \frac{k |q|}{r^2} [/tex]
 
  • #5
160
0
using the equations you specified, E = kq/r^2

for E_B = 9*10^9q_B/0.05^2 = 3.6 * 10^12 q_B

E_A = 9*10^9q_A/0.05^2 = 3.6 * 10^12 q_A

E = E_A + E_B = 3.6 * 10^12 q_B + 3.6 * 10^12 q_A
10 - 3.6 * 10^12 q_B =3.6 * 10^12 q_A
(10 - 3.6 * 10^12 q_B)/(3.6 * 10^12 ) = q_A

how do i go about finding the individual charges now?

i tried this:

i subbed in q_A = (10 - 3.6 * 10^12 q_B)/(3.6 * 10^12 ) in for q_A in the eq below

V = kq_A/r + kq_B/r, i let V = 0 and tried to solve for q_B, but the q_B's just cancelled out, what should i do now?

cheers
 
  • #6
nrqed
Science Advisor
Homework Helper
Gold Member
3,568
189
using the equations you specified, E = kq/r^2

for E_B = 9*10^9q_B/0.05^2 = 3.6 * 10^12 q_B

E_A = 9*10^9q_A/0.05^2 = 3.6 * 10^12 q_A

E = E_A + E_B = 3.6 * 10^12 q_B + 3.6 * 10^12 q_A
Be careful. You need absolute values of the charges there!
10 - 3.6 * 10^12 q_B =3.6 * 10^12 q_A
(10 - 3.6 * 10^12 q_B)/(3.6 * 10^12 ) = q_A

how do i go about finding the individual charges now?

i tried this:

i subbed in q_A = (10 - 3.6 * 10^12 q_B)/(3.6 * 10^12 ) in for q_A in the eq below

V = kq_A/r + kq_B/r, i let V = 0 and tried to solve for q_B, but the q_B's just cancelled out, what should i do now?

cheers
You know that q_A = - q_B so their absolute values are equal. Then it's easy to solve.
 
  • #7
160
0
that didn't really help, i know that their absolute values will be equal, but my unknown cancelled out and was unable to determine the charges.

where did i go wrong, what equation should i be substituting in the unknown in terms of the other unknown (q_A in terms of q_B)?

thanks
 
  • #8
dynamicsolo
Homework Helper
1,648
4
Watch your algebra -- you practically had this!

using the equations you specified, E = kq/r^2

for E_B = 9*10^9q_B/0.05^2 = 3.6 * 10^12 q_B

E_A = 9*10^9q_A/0.05^2 = 3.6 * 10^12 q_A
This is fine, so you now know that

E = E_A + E_B = 3.6 * 10^12 q_B + 3.6 * 10^12 q_A = 10 V/m ,

since the field vectors from the two charges both point in the same direction at this midpoint.

You also know, as nrqed also points out, that |q_A| = |q_B| = q , since the electric potential is zero just at this midpoint. So you can write

E = 3.6 * 10^12 q + 3.6 * 10^12 q = 10 V/m .

(The unknown should not have cancelled out. I think you misled yourself near the end there. You will get the answer you seek...)
 
  • #9
160
0
thanks, i did have that relationship earlier, i guess it was to do with how i interpreted the unknowns and their relationship.

something must of happened because using E = 3.6 * 10^12 q + 3.6 * 10^12 q = 10 V/m and solving for q, i get q = 1.39*10^12, i am supposed to get 1.39*10^-12 coulombs

what do you think happened that caused the charge to be huge instead of miniscule? i mean the equation E = kq/r^2 which would yield a huge numerator over a tiny denominator, that is understandable that it would give a big number but....

thanks
 
Last edited:
  • #10
nrqed
Science Advisor
Homework Helper
Gold Member
3,568
189
thanks, i did have that relationship earlier, i guess it was to do with how i interpreted the unknowns and their relationship.

something must of happened because using E = 3.6 * 10^12 q + 3.6 * 10^12 q = 10 V/m and solving for q, i get q = 1.39*10^12, i am supposed to get 1.39*10^-12 coulombs

what do you think happened that caused the charge to be huge instead of miniscule? i mean the equation E = kq/r^2 which would yield a huge numerator over a tiny denominator, that is understandable that it would give a big number but....

thanks
You must have made a mistake typing your numbers in the calculator. If you look at the equation it' clear that q must be of order 10^(-12) because it gets multiplied by a number of order 10^12.
 
  • #11
160
0
oh okay, thanks. that makes sense now, i got the answer i was looking for.

thanks again
 
Top