A particle with positive charge + q, is a distance r from a particle with negati

Question

A particle with positive charge + q, is a distance r from a particle with negative charge - 2q. The Coulomb force that the positive charge exerts on the negative charge is + F. where F is the magnitude and the plus sign indicates the direction of this farce. What is the magnitude and direction of the Coulomb force that the negative charge exerts on the positive charge? -F F 2F - 2F The electrostatic force that a charged particle, q1, exerts on another charged particle, Q. at a distance r1. F, A second charged particle, q2, exerts the same force F on Q at haft the distance r_1.What is the value of q_1 in relation to q2? q1 = 2 q2 q1 = 1/2 q2 q1 = 1/4 q2 q1 = 4 q2 A spherical conductor has a net positive charge. Describe how the charge is around the distributed sphere's surface. All positive charge is on the sphere surface, and all negative charge is at the center. Positive and negative charge is spread equally around the sphere's surface. The net positive charge is spread evenly throughout the sphere's volume. The net positive charge is spread evenly around the sphere's surface. How can the direction of the electric field created by an electron be correctly described? The field circulates around the electron in a plane perpendicular to the electron's spin. The field points away from and towards the electron. The field points away from the electron. The field points towards the electron.

Explanation / Answer

1. The magnitude will be same, just the direction will be reversed, i.e (-F).

2. The relation between coulombic force,charges and distance between the 2 charges is F= kq1q2/r2. K being consant.

So equating force due to both the charges : kq1Q/r2 = k q2Q/(r/2)2  

So, q1 =4* q2

3. The net positive charge is spread evenly around the surface of the sphere. No charge is inside the solid conductor. (Faraday cage)

4. Electron has negative charge and the electric field is always directed towards the negative charge. It is away from the charge if charge is positive.

This is considered by the direction of the positive test charge in the field of that negative charge. So (d) is correct.

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