Electric Force Calculation Using Coulomb's Law | Problem 3, 9

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In summary: F = qE cos(theta) where q = charge, E = electric field, and theta is the angle between the field and the direction of the dipole. In summary, the electric force between the two charges is 190.1888733N.
  • #1
mustang
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Problem 3.
Two identical conduction spheres are placed with their centers 0.34m apart. One is given a charge of +13*10^-9 C amd the other is given a charge of -14*10^-9 C. The spheres are connected by a conductiong wire.
After equilbrium has occured, find the electric force between the teo spheres. Answer in N.
Note: Would you add up the charges divide by two and the put it in Colulomb's law were it is the charged squared? From there find the electric force?

Problem 9.
Three point charges , q_1=+6.6 uC, q_2=+2.4uC, and q_3=-2.4uC, lie along the x-axis at x=ocm, x-2.4 cm, and x=5.3 cm, respectively.
a. What is the force exerted on q_1 by the two charges? (To right is positive) Use 8.99*10^9 N*m^2/C^2. Answer in N.
Note: How would put these values in Coulomb's law?
 
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  • #2
yes for first part

Find the field at the point q1 due to q2 & q3 sayit is E

Then force will be q1E
 
  • #3
re #9:

remember that E is a vector. THe "+" and "-" values that you get with Coulomb's equation do not tell you the direction is to the right or to the left. Draw a picture for E from q2 and from q3. IF they point in the same direction then the two E's add up. IF they point in different directions, then you subtract values, and the direction of the net E is the same as the larger E.
 
  • #4
From Couloumbs law u can calculate the direction

the Vector form for Coulombs law is

[tex]
\vec F = \frac{kq_1q_2}{\vec r^3} \vec r
[/tex]

The thing is u have to put the sign of q1&q2 along with magnitude
 
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  • #5
regarding question 9

To find magnitiude of the force of q_1 don't I use Coulomb's law twice. Once with the second charge to be 2.4uC and another time with second charge -2.7uC, with both charges being multiplyied with 6.6uC.
Then I will get two amounts whih i would multiply with what?
 
  • #6
I was just giving the formula in vector form

U got it right u will have to do with charges

Got the force due to both charges say F1,F2

Apply the superposition Net force= F1+F2

Note you have to add vectors force is vector
 
  • #7
Is the answer -190.1888733N.
 
  • #8
U never get such an enormous force generally in Coulombs law or in Electrostatics

I just noticed there is a simple way of doing the problem

The given set of charges forms a dipole

U can use the equation of Field in axial line for a dipole
 

What is Coulomb's Law?

Coulomb's Law is a fundamental law in physics that describes the electrostatic interaction between two charged particles. It states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

How is Coulomb's Law used to calculate electric force?

Coulomb's Law is used to calculate the magnitude of the electric force between two charged particles. The formula for calculating electric force is F = k * (q1 * q2) / r2, where F is the force, k is Coulomb's constant, q1 and q2 are the charges of the particles, and r is the distance between them.

What is the unit of electric force?

The unit of electric force is Newton (N). This is the same unit used to measure other types of forces, such as gravitational force.

What are the potential consequences of not considering the direction of the force in Coulomb's Law calculations?

If the direction of the force is not considered in Coulomb's Law calculations, the calculated result may not accurately represent the actual force between the two charged particles. This can lead to incorrect predictions and inaccurate understanding of electrostatic interactions.

Can Coulomb's Law be applied to all types of charged particles?

Yes, Coulomb's Law can be applied to all types of charged particles, including both point charges and charged objects with finite sizes. However, for charged objects with finite sizes, the calculation becomes more complex and involves integration.

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