Finding charges using electric field and distance

You can solve for q_A or q_B and then substitute that value into the other equation to solve for the other charge. Alternatively, you can use the relationship that q_A = -q_B and solve for the magnitude of one charge, then substitute that into the other equation to find the magnitude of the other charge.
  • #1
scholio
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 into 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:
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  • #2
scholio said:

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 into 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
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?
 
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  • #4
scholio said:
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
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 canceled out, what should i do now?

cheers
 
  • #6
scholio said:
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 canceled 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
that didn't really help, i know that their absolute values will be equal, but my unknown canceled 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
Watch your algebra -- you practically had this!

scholio said:
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 canceled out. I think you misled yourself near the end there. You will get the answer you seek...)
 
  • #9
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
 
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  • #10
scholio said:
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
oh okay, thanks. that makes sense now, i got the answer i was looking for.

thanks again
 

1. What is an electric field?

An electric field is a region in space where an electric charge experiences a force. It is a vector quantity, meaning it has both magnitude and direction.

2. How do you find the electric field at a given point?

The electric field at a given point can be found by dividing the force experienced by a test charge at that point by the magnitude of the test charge. This can be expressed as E = F/q, where E is the electric field, F is the force, and q is the test charge.

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

As distance increases, the electric field decreases. This is because the force between two charges is inversely proportional to the square of the distance between them. Thus, as distance increases, the force and electric field weaken.

4. How can electric field be used to find charges?

By measuring the electric field at a given distance from a charge, the charge can be calculated using the equation E = kq/r^2, where k is the Coulomb constant, q is the charge, and r is the distance. Solving for q gives us q = Er^2/k.

5. What are some real-world applications of finding charges using electric field and distance?

One real-world application is in particle accelerators, where charged particles are accelerated using electric fields. Another application is in capacitors, where the electric field is used to store energy. Electric field and distance calculations are also important in understanding the behavior of lightning and designing lightning protection systems.

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