Physics A: Problem Set 3: Electric Field
recommended reading
Barron's Let's Review:  8.7–8.9 The Electric Field 
physics.info:  Electric Field 
Wikipedia:  Electric field 
HyperPhysics:  Electric Field 
Khan Academy:  Electrostatics (part 2) 
drawings in class
 Sketch the electric field around the following arrangements of "small" charged objects.
 an isolated positive charge
 an isolated negative charge
 two positive charges of equal magnitude
 two negative charges of equal magnitude
 an electric dipole (one positive and one negative charge of equal magnitude)
drawings and sentences for classwork
 Sketch the electric field around the following pairs of point charges. Draw continuous field lines and assume the charges are separated by a few centimeters of empty space.
 A +3 μC charge on the left and a +1 μC charge on the right.
 A +3 μC charge on the left and a −1 μC charge on the right.
The bigger charge needs to dominate the field diagram. Ideally, one would like three times more field lines coming out of the larger charge than go into or come out of (depending on the sign of the charge) the smaller charge. You should make drawings that look something like these from flashphysics.org…
or these from academo.org…
 A charge of −1.0 μC is located on the yaxis 1.0 m from the origin at the coordinates (0,1) while a second charge of +1.0 μC is located on the xaxis 1.0 m from the origin at the coordinates (1,0). Determine the direction of the electric field at the origin.
Imagine a small positive test charge placed at the origin. Where does it want to go? The negative charge above pulls the imaginary test charge up, since opposite charges attract. The positive charge on the right pushes the imaginary test charge left, since like charges repel. The field at the origin is therefore a combination of up and to the left. Since the two charges are of equal size and equal distance from the origin, both are contributing equally to the electric field. The resultant electric field is then the hypoteneuse of a right triangle with two equal legs — a 45°, 45°, 90° triangle in the second quadrant. This is 135° as a standard angle or 45° above the −x axis.
 The electric field of the Earth is due to the separation of charges between the surface of the Earth and the upper layers of the Earth's atmosphere.
 If the direction of the Earth's electric field points down, what is the sign of the charge on the Earth's surface? Explain your answer.
 Is the charge on the surface of the Earth due to an excess or a deficit of electrons? Explain your answer.
 An estimate of the net charge on the Earth can be made by assuming that all of the charge on the Earth is concentrated at its center. If the electric field on the Earth's surface is 120 N/C, what is the net charge of the Earth?
 Calculate the surface area of the Earth.
 What is the surface charge density of the Earth in…
 coulombs per square meter [C/m^{2}] and
 elementary charges per square meter [e/m^{2}].
The Earth's surface is slightly negative since electric field lines converge on (or end on) negative charges. Downward pointing field lines are converging on the Earth's surface.
The charge on the Earth's surface is due to an excess of electrons since electrons carry a negative charge.
Use the equation for the electric field surrounding a point charge.
E = k q r^{2} Solve it for charge.
q = Er^{2} k Use the radius of the Earth as the distance to the charge.
q = (120V/m)(6.37 × 10^{6}m)^{2}) (8.99 × 10^{9} Nm/C^{2}) Compute.
q = 5.42 × 10^{5} C
Use the equation for the surface area of a sphere with the Earth's radius.
A = 4πr^{2} A = 4π(6.37 × 10^{6}m)^{2} Surface charge density (σ, sigma) is the ratio of charge (q) to area (A).
σ = q A Do it once using coulombs.
σ = 5.42 × 10^{5} C = 1.06 × 10^{−9} C/m^{2} 5.10 × 10^{14} m^{2} Then convert that to elementary charges.
σ = 1.06 × 10^{−9} C/m^{2} = 6.63 × 10^{9} e/m^{2} 1.60 × 10^{−19} C/e
computation for homework

The more common unit for the electric field is the volt per meter [V/m]. An old fashioned television picture tube (also known as a cathode ray tube or CRT) is about a half meter long and has a potential difference of about 10,000 V between the front and the back. Determine the following quantities inside a typical CRT TV.
 the electric field strength
 the force on an electron
 the acceleration of an electron
 its velocity
 its kinetic energy
 Are the xrays produced by a CRT television a matter for concern? If yes, why? If not, why not?
The electric field strength can be determined just by looking at the units (a process formally called dimensional analysis). The problem says electric field is often stated in units of volts per meter. Take the volts (the voltage or electric potential across the tube) and divide by the meters (the length of the tube).
E = V d E = 10,000 V 0.50 m E = 20,000 V/m Just look at the units again to find the force on an electron inside the tube. The other SI unit for electric field is the newton (force) per coulomb (charge), which is the the same as the volt (potential) per meter (distance). Look up the charge on the electron in the reference source of your choice.
E = F q 20,000 N/C = F 1.6 × 10^{−19} C F = 3.2 × 10^{−15} N The acceleration of an electron in the tube is found using Newton's second law of motion. Look up the mass of the electron in a reference source. Electrons are tiny little things, so they accelerate easily.
a = F m a = 3.2 × 10^{−15} N 9.1 × 10^{−31} N a = 3.5 × 10^{15} m/s^{2} Find the velocity at the end of the tube using the acceleration we just calculated and an appropriate equation of motion. May I suggest this one. (Remember, we're starting from rest in this problem.)
v^{2} = v_{0}^{2} + 2a∆s v = √[2(3.5 × 10^{15} m/s^{2})(0.5 m)] Need to know the kinetic energy of the electron? There's an equation for that.
K = ½mv^{2}
K = ½(9.1 × 10^{−31} kg)(59,000,000 m/s)^{2}
K = 1.6 × 10^{−15} JThe glass in the screen of a CRT contains lead, which absorbs xrays. You do not need to be concerned about the the xrays produced by a CRT television (especially if you don't have one).