Electrostatic Force on b: Positive & Negative?

In summary: Yes, the force on b due to c is "pointing up", and the force on b due to a is "pointing down", and the magnitudes of those forces are equal (from your equation). And so the vector sum of all the forces on b is zero in the situation shown. However, if you take the square of the distance between the charges, then the force on b due to c will be greater than the force on b due to a (because the distance between the charges has increased). So the force on b due to c will be equal to 2F. Yes, the force on b due to c is "pointing up", and the force on b due to a is "pointing down",
  • #1
Air
203
0
1. All variables and given/known data:

Diagram-1.jpg

[itex]a = c = 1 \times 10^{-19} C[/itex]
[itex]b = 2 \times 10^{-19} C[/itex]
Distance between a&b = b&c = 0.1m (b at origin, a at (0, 0.1) and c at (0, -0.1))


2. Homework Equations :

[itex]F = k\frac{|q_1||q_2|}{r^2}[/itex]


3. The problem that I'm having:

Am I correct to think that the electrostatic force on b from other charges is at a minimum because the Force will be positive and negative (due to the nature of r ([itex]\pm[/itex] 0.1)) cancelling each other out? :confused:
 
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  • #2
Air said:
1. All variables and given/known data:

Diagram-1.jpg

[itex]a = c = 1 \times 10^{-19} C[/itex]
[itex]b = 2 \times 10^{-19} C[/itex]
Distance between a&b = b&c = 0.1m (b at origin, a at (0, 0.1) and c at (0, -0.1))


2. Homework Equations :

[itex]F = k\frac{|q_1||q_2|}{r^2}[/itex]


3. The problem that I'm having:

Am I correct to think that the electrostatic force on b from other charges is at a minimum because the Force will be positive and negative (due to the nature of r ([itex]\pm[/itex] 0.1)) cancelling each other out? :confused:

Yes, the force on b due to a is "pointing down", and the force on b due to c is "pointing up", and the magnitudes of those forces are equal (from your equation). And so the vector sum of all the forces on b is zero in the situation shown.
 
  • #3
olgranpappy said:
Yes, the force on b due to a is "pointing down", and the force on b due to c is "pointing up", and the magnitudes of those forces are equal (from your equation). And so the vector sum of all the forces on b is zero in the situation shown.

I just realized that the radius will be squared hence the negative will also produce a positive force. How does it equal zero then (Surely, it will just equal 2F)? Am I to use another formula or is the electrostatic formula correct to use?
 

1. What is electrostatic force?

Electrostatic force is the force of attraction or repulsion between electrically charged objects.

2. How does electrostatic force act on positive and negative charges?

The electrostatic force acts between positive and negative charges in opposite directions. Positive charges attract negative charges, while positive charges repel other positive charges and negative charges repel other negative charges.

3. What is the formula for calculating electrostatic force?

The formula for calculating electrostatic force is F = k * (q1 * q2)/r^2, where k is the Coulomb's constant, q1 and q2 are the charges of the two objects, and r is the distance between them.

4. How does distance affect the electrostatic force between two charges?

The electrostatic force between two charges is inversely proportional to the square of the distance between them. This means that as the distance increases, the force decreases.

5. Can objects with the same charge have an electrostatic force between them?

Yes, objects with the same charge can have an electrostatic force between them. However, this force will be repulsive, as objects with the same charge repel each other.

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