Coulomb's Law and Electric Field
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Coulomb's Law


We want to distribute a charge Q between two bodies. One of the bodies receives a charge q1 and the other a charge q2. The distribution of the charges is done in such a way that q1+q2=Q. Determine the ratio between the charges so that the Coulomb's force of repulsion between q1 and q2 is maximum for any distance between the charges.

Two charges of the same magnitude and opposite signs are fixed on a horizontal line at a distance d from each other. A sphere, with mass m carries an electric charge, attached to a wire and is approximate, first of one of the charges until it is in equilibrium exactly on it at a height d. Then, the wire is moved toward the second charge until the charge is in equilibrium over the second charge. Determine the angles of deviation in both cases, knowing that in the first case, the deviation angle is twice higher than the deviation angle in the second case.

Electric Field of a Discrete Charge Distribution


Two identical charges of the same sign are separated by a distance of 2d. Calculate the electric field vector at points along the perpendicular bisector of the line joining the two charges. Verify the solution for points far away from the center of the system.
Solution

Suggestion: compare with the electric field obtained as a scalar.

Two equal charges of the same sign are separated by a distance of 2d. Calculate the electric field vector at the points along the perpendicular bisector of the line joining the two charges. Check the solution for points far from the charges.
Solution

Suggestion: compare with the calculation of the  magnitude of the electric field

Two identical charges of the same sign are separated by a distance of 2d. The magnitude of the electric field at the points along the perpendicular bisector of the line joining the two charges is given by
\[ \begin{gather} E=\frac{1}{4\pi \epsilon_0}\frac{2qy}{\left(a^2+y^2\right)^{3/2}} \end{gather} \]

Determine:
a) The points along the y-axis, for which the magnitude of the electric field assumes its maximum value;
b) The magnitude of the maximum electric field.
Solution

Suggestion: compare with the maximum points of the  electric field of a loop charged with charge Q.

Electric Field of a Continuous Charge Distribution


A ring of radius a carries a uniformly distributed electric charge Q. Calculate the electric field vector at a point P on the symmetry axis perpendicular to the plane of the ring at a distance z from its center.

A ring of radius a is uniformly charged with a charge Q. The electric field produced by this ring at points on the axis of symmetry perpendicular to the plane of the ring at a distance z is given, in magnitude, by
\[ \begin{gather} E=\frac{1}{4\pi \epsilon_{0}}\frac{Qz}{\left(a^{2}+z^{2}\right)^{3/2}} \end{gather} \]

Determine:
a) For what values of z is the electric field maximum?
b) What is this maximum value.

Two concentric rings are located on the same plane. The ring of radius R1 has a charge Q1, and the ring of radius R2 has a charge Q2. The electric field vector produced by a ring of radius r at a distance z from the center is given by
\[ \begin{gather} \mathbf{E}=\frac{1}{4\pi \epsilon_0}\frac{Qz}{\left(r^2+z^2\right)^{3/2}}\;\mathbf{k} \end{gather} \]

Determine the electric field vector:
a) In the common center of the two rings;
b) At a point located at a distance z, much greater than R1 and R2.

An arc of a circle of radius a and central angle θ0 carries an electric charge Q uniformly distributed along the arc. Determine:
a) The electric field vector, at the points of the line passing through the center of the arc and is perpendicular to the plane containing the arc;
b) The electric field vector in the center of curvature of the arc;
c) The electric field vector when the central angle tends to zero.

A ring of radius a carries a uniformly distributed electric charge q1 in one of the halves and q2 in another half. Calculate the electric field vector at a point P on the symmetry axis perpendicular to the plane of the ring at a distance z from its center.
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