A +'ve charge Q is fixed to the origin. A -'ve charge 9Q

In summary: That is, F1+F2=0. Solve the two equations for the value of x.In summary, the problem requires finding the point where a small positive test charge will experience zero electrical force from a positive charge and a negative charge with 9 times its magnitude positioned a distance d to the right of the origin on the positive x-axis. This can be solved by writing equations for the forces on the test charge from each of the two charges and setting their sum equal to zero, then solving for the value of x.
  • #1
abm77
13
0

Homework Statement



A positive charge Q is fixed to the origin. A negative charge with magnitude 9Q is fixed along the positive x-axis a distance d to the right of the origin. Determine a point where a small positive test charge q will experience zero electrical force from the two charges, or show that there is no such point. Draw diagrams to help you.

Homework Equations


[/B]
FE = kq1q2 / r2

The Attempt at a Solution



(+Q)--------d---------(-9Q)

I know the test charge must be to the left of the positive charge Q, and I tried to solve by rearranging the equation to solve to r, but that leaves you with dividing the force, which in this case is zero which does not work.

r = √kq1q2 / FE
But FE = 0 so this does not work.
 
Physics news on Phys.org
  • #2
Your equation is for the force between two charges. There are three in the problem. The net force must sum to zero not the two individual forces.
 
  • #3
CWatters said:
Your equation is for the force between two charges. There are three in the problem. The net force must sum to zero not the two individual forces.
So would the equation for three charges just include a q3 at the end? Thus making the small positive test charge 8Q to have it equal zero?
 
  • #4
abm77 said:
So would the equation for three charges just include a q3 at the end?

Not sure I follow that.

The test charge q is +ve so there are two forces on it..

1) A repulsive force between +Q an +q
2) An attractive force between -9Q and +q

The question asks where must +q be for these two forces sum to zero. Have you studied vector addition?
 
  • #5
CWatters said:
Not sure I follow that.

The test charge q is +ve so there are two forces on it..

1) A repulsive force between +Q an +q
2) An attractive force between -9Q and +q

The question asks where must +q be for these two forces sum to zero. Have you studied vector addition?

How would vector addition work for this question as there are no values for the variables?
 
  • #6
abm77 said:
How would vector addition work for this question as there are no values for the variables?
The values are irrelevant, only the relative size of the charges is important (and you know that one has 9 times more charge than the other).

Place the test charge at an arbitrary position on the x-axis, then write an equation for F1 the force on the test charge from the one charge and another equation for F2 the force on the test charge from the other charge. The net force is F1+F2. The problem asks you to find where this net force is zero.
 
  • Like
Likes CWatters

FAQ: A +'ve charge Q is fixed to the origin. A -'ve charge 9Q

1. What is the force between the two charges?

The force between two charges is given by Coulomb's law, which states that the force is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them. In this case, the force would be attractive, as opposite charges attract each other.

2. How does the force change if the distance between the charges is doubled?

If the distance between the two charges is doubled, the force between them will decrease by a factor of four. This is because the force is inversely proportional to the square of the distance. So, if the distance is doubled, the force will be one-fourth of its original value.

3. What is the direction of the force on each charge?

The direction of the force on each charge can be determined by the type of charges. Since one charge is positive and the other is negative, the force will be attractive and the direction of the force on the positive charge will be towards the negative charge, and vice versa.

4. How does the force change if the magnitude of the negative charge is doubled?

If the magnitude of the negative charge is doubled, the force between the charges will also double. This is because the force is directly proportional to the product of the two charges. So, if one charge is doubled, the force will also double.

5. What would happen if the charges were of the same magnitude?

If the charges were of the same magnitude, they would repel each other instead of attracting. This is because like charges repel each other, while opposite charges attract. The force between the charges would still be given by Coulomb's law, but the force would be repulsive instead of attractive.

Back
Top