To understand Coulomb\'s law, electric fields, and the connection between the el

Question

To understand Coulomb's law, electric fields, and the connection between the electric field and the electric force.

Coulomb's law gives the electrostatic force F?  acting between two charges. The magnitude F of the force between two charges q1 and q2 depends on the product of the charges and the square of the distance r between the charges:

F=k|q1q2|r2,

where k=1/(4??0)=8.99×109N?m2/C2. The direction of the force is along the line connecting the two charges. If the charges have the same sign, the force will be repulsive. If the charges have opposite signs, the force will be attractive. In other words, opposite charges attract and like charges repel.

Because the charges are not in contact with each other, there must be an intermediate mechanism to cause the force. This mechanism is the electric field. The electric field at any location is equal to the force per unit charge experienced by a charge placed at that location. In other words, if a charge q experiences a force F? , the electric field E?  at that point is

E? =F? q.

The electric field vector has the same direction as the force vector on a positive charge and the opposite direction to that of the force vector on a negative charge.

If the total positive charge is Q = 1.62×10?6 C , what is the magnitude of the electric field caused by this charge at point P, a distance d = 1.53 m from the charge? (Figure 1)

Enter your answer numerically in newtons per coulomb.

Explanation / Answer

Electric field due to a point charge is E = k*Q/d^2

k = 8.99*10^9 N-m^2/C^2

Q = 1.62*10^-6 C

d = 1.53 m

E = k*Q/d^2 = (8.99*10^9*1.62*10^-6)/(1.53^2)

E = 6221.5 N/C

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