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_vec acting between two charges. the magnitude f of the force between two charges q_1 and q_2 depends on the product of the charges and the square of the distance r between the charges:
f= k\frac{|q_1\,q_2|}{r^2},
where k = 1/(4\pi\epsilon_0)= 8.99 \times 10^9 \; {\rm n\cdot m^2/c^2}. 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_vec, the electric field e_vec at that point is
{\vec e}=\frac{\vec 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? (part c figure)
enter your answer numerically in newtons per coulomb.