Electric Charge and Circuits Exam Notes
Electric Charge
Law of Charges: Like charges repel, unlike charges attract.
Electrons transfer when fur is rubbed against a rubber rod, giving rubber a net negative charge and fur a net positive charge.
Electrons transfer from a glass rod to silk when rubbed, leaving the glass rod with a net positive charge and the silk with a net negative charge.
Charge magnitude of electrons and protons is equal to the elementary charge, e:
e = 1.60 \times 10^{-19} \text{ coulombs (C)}
e = elementary charge = smallest charge measured on an isolated particle.
Coulombs (C) are the SI unit for charge.
Charge on electron = -e, charge on proton = +e.
Protons are much more massive than electrons.
Electrons are elementary particles; protons are not (they are composed of quarks).
Quarks are elementary particles composing protons and neutrons.
q_{up \ quark} = + \frac{2}{3}e
q_{down \ quark} = - \frac{1}{
q proton = 2qup quark + 1q down quark = 2(+2/3e)+(-1/3e)=+4/3-1/3e=+e
q{neutron} = 1q{up \ quark} + 2q_{down \ quark} = (+\frac{2}{3}e) + 2(-\frac{1}{3}e) = 0
Interactions of a Rubber Balloon with Hair
Balloon Sticking to Hair:
Electrons transfer from hair to balloon when rubbed.
Balloon becomes negatively charged, and hair becomes positively charged.
Unlike charges attract, pulling hair and balloon together.
Hair Strands Repelling:
Electrons transfer from hair to balloon.
Hair becomes positively charged.
Like charges repel, pushing hair strands apart.
Masses of Electron and Proton
m_{electron} = 9.11 \times 10^{-31} \text{kg}
m_{proton} = 1.67 \times 10^{-27} \text{kg}
\frac{m{proton}}{m{electron}} = \frac{1.67 \times 10^{-27} \text{kg}}{9.11 \times 10^{-31} \text{kg}} \approx 1830
Polarization
Polarization, used when the balloon sticks to a wall, will be discussed in a future lesson.
Quantization of Charge
Elementary charge is very small.
Example: Calculating the number of excess protons needed for a charge of 1 coulomb.
q = ne
q = net charge on an object.
n = excess number of charge carriers.
e = elementary charge.
Rearranging the formula to find the number of protons:
n = \frac{q}{e} = \frac{1 \text{C}}{1.60 \times 10^{-19} \text{C/proton}} = 6.25 \times 10^{18} \text{ protons}
Example: Determining if an object can have a net negative charge of 2.00 \times 10^{-19} coulombs.
n = \frac{q}{e} = \frac{-2.00 \times 10^{-19} \text{C}}{-1.60 \times 10^{-19} \text{C/electron}} = 1.25 \text{ electrons}
Object cannot have a net negative charge of 2.00 \times 10^{-19} coulombs.
Charge is quantized: it comes in discrete quantities that are multiples of the elementary charge.
Charge is caused by having more or fewer charged particles (protons or electrons).
The charge on an object, q, must be an integer multiple of the elementary charge, e.
In q = ne, n must be an integer, because you cannot cut protons and electrons into pieces.
Coulomb's Law or the Electric Force
Coulomb’s Law describes the magnitude of the electric force between two charged particles.
Fe=kq1q2/r²
F_e = electric force.
k = Coulomb constant (8.99 \times 10^9 \frac{N \cdot m^2}{C^2}).
q1 & q2 = electric charges.
r = distance between the centers of charge.
Similarities to Newton’s Universal Law of Gravitation:
Fg = Gm1m2/r²
The Coulomb constant, k, is much larger than the Universal Gravitational Constant, G.
\frac{k}{G} = \frac{8.99 \times 10^9 \frac{N \cdot m^2}{C^2}}{6.67 \times 10^{-11} \frac{N \cdot m^2}{kg^2}} = 1.347826087 \times 10^{20} \approx 1.35 \times 10^{20} \frac{kg^2}{C^2}
A point charge is like a point mass, but relates to charge instead of mass. A point charge has zero size and carries an electric charge, or its mass is negligible compared to its charge.
Prefixes for coulombs:
Micro (µ): 1 \mu C = 1 \times 10^{-6} C
Nano (n): 1 nC = 1 \times 10^{-9} C
Pico (p): 1 pC = 1 \times 10^{-12} C
Example 1
Find the electric force btwn two equal magnitude point charges 2.0m apart, charges are both 5.0 \mu C, one is positive/negative.
q_1 = +5.0 \mu C = 5 \times 10^{-6} C
q_2 = -5.0 \mu C = -5 \times 10^{-6} C
r = 2.0 m
Fe = k \frac{q1 q_2}{r^2} = (8.99 \times 10^9) \frac{(5 \times 10^{-6})(-5 \times 10^{-6})}{2^2} = -0.0561875 \approx -0.056 N
Each charge experiences approximately 0.056 N of force toward the other charge.
Coulomb's Law Versions
Fe = k \frac{q1 q_2}{r^2}
Magnitude of the electric force (ignoring direction): Fe = k \frac{|q1 q_2|}{r^2}
Unit vector version: \vec{F{12}} = k \frac{q1 q2}{r{12}^2} \hat{r}_{12}
\vec{F{12}} = the electric force by charge 1 on charge 2; \hat{r}{12} = unit vector directed from charge 1 toward charge 2.
Using the first version:
A negative force: attractive force.
A positive force: repulsive force.
Fg is always attractive, but Fe can be attractive or repulsive.
Building on Example #1 (Example #2)
If a third charge, q3, with the same charge as the positively charged object (q1), is placed 2.0 meters on the opposite side from the negative charge (q2), what is the net force acting on q1 (the positive charge in the middle)?
Adding the third charge does not affect the electric force between q2 and q1, which is 0.056 N acting to the right.
q1 and q3 both positive, force 3-1 is positive, repulsive force, on q_1 that would be to the right.
F{31} = k \frac{q1 q3}{r{31}^2} = (8.99 \times 10^9) \frac{(+5 \times 10^{-6})(+5 \times 10^{-6})}{2^2} = 0.0561875 N
\sum F{on1} = F{21} + F_{31} = 0.0561875 + 0.0561875 = 0.112375 \approx 0.11 N
Charging via Conduction
Charge the balloon by rubbing fur on it (electrons move from fur to rubber).
Bring the charged balloon close to the electroscope and foils move apart
Touch the balloon to electroscope so that the electroscope is charged by conduction
Touch electroscope and foils fall down to original positions
Charging by Friction: Rubbing fur on balloon causes electrons to move from the fur to the rubber balloon. The fur now has excess positive charge and the balloon has excess negative charge.
The electroscope has an equal number of protons and electrons, net charge electroscope = 0.
Balloon has larger number of electrons then protons. The negatively charged balloon is brought close to the electroscope, however, the electroscope and the balloon have not touched yet. The aluminum foils are pushed away from one another.
The foils have the same charge. The foils have an electric force pushing them apart.
Electrons in the electroscope, which are negatively charged, are repelled by the negative charges in the balloon and flow down to the metal foils. The metal foils have an excess of electrons, or a net negative charge, and an electric force pushes the metal foils away from one another.
The top of the electroscope has a net positive charge because there are fewer electrons than protons in the ball at the top of the electroscope. The electroscope still has a net neutral charge.
Touching: The balloon touches the electroscope, electrons transfer from the balloon to the electroscope. The balloon and electroscope now both have excess electrons, however, total number of excess electrons in the system remains the same (Conservation of Charge). Because the balloon and electroscope have excess electrons, the foils also have excess electrons and repel one another.
Touching electroscope with a finger: Touching electroscope grounds the electroscope resulting in in the electroscope to no longer be electrically isolated and electrons transfer from electroscope into ground, the electroscope has neutral charge.
An Ideal Ground is an infinite well of charge carriers. Electrical circuits are literally connected to the Earth or the “ground”. The Earth has an infinite number of electrons which we can pull from it, or we can give to it.
If something goes wrong in a circuit, the “ground” will serve as a way to balance out the charges.
Touching the electroscope grounds it and excess electrons flow from the electroscope into the “ground” and the electroscope is now electrically neutral. 2 items to remember about Charging by Conduction:
The two objects have to touch (The balloon and the electroscope).
The two objects end with the same sign of net charge (Both end with an excess negative charge. Before the electroscope is grounded and ends with a net neutral charge).
Charging by Induction
charge the balloon
bring balloon close to electroscope
ground the electroscope
remove the ground
remove the balloon and electroscope is charged by induction
ground the electroscope
When the electroscope is grounded the negatively charged balloon is held near the electroscope, therefore, electrons in the electroscope flow into the ground (Law of Charges, like charges repel one another). The ground essentially provides an escape route
for electrons to leave the electroscope. The ground essentially provides an escape route for electrons to leave the electroscope.Electrons still in the electroscope, just fewer than before, and many are in the metal foils of the electroscope because they are repelled from the electrons in the balloon, that gives the metal foils a net neutral charge and the two foils are not repelled from one another.
balloon removed, electrons in the electroscope are repelled from one another in the electroscope and get distributed throughout the electroscope, which leaves the electroscope with an excess of protons and a net positive charge (metal foils are repelled from one another, because the positive charges in the foils repel one another).
electroscope is grounded means electrons are pulled from the ground into the electroscope to balance out the excess protons in the electroscope and leave the electroscope with an equal number of protons and electrons and a net neutral charge.
2 items to remember about Charging by Induction:
The two objects do not have to touch.
The two objects end with opposite sign of net charge, the electroscope ends with an
excess of positive charge and the balloon ends with an excess negative charge (Before the electroscope is grounded and ends with a net neutral charge).
Polarization of Charge
A charged balloon is attracted to a wall because the molecules in the wall become polarized.
Polarization does not mean the wall becomes charged, it simply means the molecules in the wall have aligned themselves such that there will be a net attractive force between the wall and the charged balloon.
electrons in the wall move away from the electrons in the balloon (Law of Charges, the electrons in the wall will be repelled from electrons in the balloon because they both have negative charges/like charges repel and unlike charges attract).
all the like charges are farther apart than all of the unlike charges.
Coulomb's Law: the closer the charged objects are to one another, the smaller the “r” value in Coulomb’s Law and therefore the larger the electric force.
the opposite charges are closer than the like charges, the attractive electric force is larger than the repulsive electric force and the net electric force between the balloon and the wall is an attractive force (a charged object can be attracted to a neutrally charged object).
attracting by polarization (charged balloon can pick up little pieces of paper & cause an aluminum can to roll)
The electric forces in the polarization demonstrations are quite small (masses of the balloon and little pieces of paper are small, so only a small electric force is required to hold them up and it only requires a small force to roll the aluminum can).
The electric force caused by polarization is typically larger for a conductor than an insulator (in insulators, electrons are just pushed to the opposite side of the atom, however, in conductors, the
electrons are free to move about more and actually end up farther away-> larger difference in attractive vs repulsive force and a larger net electric force in a conductor than an insulator).
Conservation of Charge
The total electric charge of an isolated system never changes. Isolated system: the universe.
Add up all the positive charges and subtract all the negative charges and you will always get the same number or conductive metal pieces of an electroscope which are electrically isolated from the rest of the universe by the rubber and glass insulators (the net electric charge of the electroscope will remain constant, as long as it remains isolated).
Example Problem #1
Two charged, conducting objects collide and separate. Before colliding, the charges on the two objects are +3e and -6e. Which of the following are possible values for the final charges on the two objects? Choose all possible answers. (a) +4e, -7e (b) +2e, -2e (c) -1.5e, -1.5e (d) -3.5e, +2.5e (e) +e, -4e
a works b, c ,d, doesn't work e works.
Correct answers are (a) and (e) because those are the only two options which have a total final charge equal to the total initial charge and are integer multiples of the fundamental charge e.
Example Problem #2
Two identical, conducting spheres are held using insulating gloves a distance x apart. Initially the charges on each sphere are +3.0 pC and +6.0 pC. The two spheres are touched together and returned to the same distance x apart. You may assume x is the distance between their centers of charge.
(a) What is the final charge on each sphere?
(b) Is the final electric force between the two spheres increased, decreased, or the same when compared to the initial electric force?
Because the two spheres are identical, after touching, the spheres will have equal charge.
(a) Both charges end with 4.5 pC of charge.
This is 4.5 picocoulombs of charge or 4.5 x 10-9 C which an object is physically able to have (Imagine that: 28 billion more protons than electrons on each sphere. Each sphere has a deficit of 28 billion electrons and therefore has a net charge of 4.5 pC. )
The two spheres have like charges, so they are repelled from one another with an electric force with a magnitude of.
Therefore Final electric force is greater than the initial electric force
Bohr Model of the Hydrogen Atom
Determine the speed of the electron. The electron orbits at a radius of 5.29 \times 10^{-11}\text{ m}
Free body diagram of the electron shows one force, the electric force, inward toward the proton.
Balloon Excess Charges Experiment
Two 0.0018 kg balloons each have approximately equal magnitude excess charges and hang as shown. If \theta = 21° and L = 0.39 \text{m}, what is the average
number of excess charges on each balloon?Free Body Diagram (left balloon).
Break Force of Tension into its components.
Redraw the Free Body Diagram.
Electric force is roughly 2 million million times larger than the force of gravity so the force of gravity is negligible.
what is the average number of excess charges on each balloon?
Electric Fields
If we were to place a positive test charge in an electric field, it would experience an electrostatic force. An electric field is the ratio of the electrostatic force the test charge would experience and the charge of the test charge.
A positive test charge :a charge which is small enough not to measurably change the electric field it is placed in. Electric field directions are defined according to the direction of the net electrostatic force on a positively charged test charge.
The equation for an electric field and its units:
E = \frac{F_e}{q}
Electric field and the electrostatic force experienced by a positive charge in the electric field will be in the same direction
the electric field which surrounds a point charge:
E = k \frac{q}{r^2}
The gravitational field around a planet has a similar format:
g = G \frac{m}{r^2}
The electric field which exists between two large, parallel, oppositely charged plates is similar to the gravitational field close to the surface of a planet:
Electric field is a vector which means that the electric field for two point charges which are near one another is the sum of the two individual electric fields for each point charge.
Electric field maps like the one above are simplified models and vector maps which show the magnitude and direction of the electric field for the entire region.
Electric Field Lines Basics:
In the direction a small, positive, test charge would experience an electrostatic force
Electric field lines per unit area is proportional to electric field strength (Higher density electric field lines = higher electric field strength)
Start on a positive charge and end on a negative charge (or infinitely far away if more of one charge than another)
Start perpendicular to the surface of the charge. Electric field lines never cross
Continuous Charge Distributions
Continuous Charge Distribution: A charge that is not a point charge having shape and continuous charge distributed throughout the object.
In order to find the electric field which exists around a continuous charge distribution, we can use Coulomb’s Law and the equation definition of an electric field.
We consider the charged object to be made up of an infinite number of infinitesimally small point charges dq and add up the infinite number of electric fields via superposition is an integral.
Realize that, students are only expected to be able to use this equation to determine electric fields around continuous charge distributions with high symmetry.an infinitely long, uniformly charged wire or cylinder at a distance from its central axis
a thin ring of charge at a location along the axis of the ring
a semicircular arc or part of a semicircular arc at its center
a finite wire or line of charges at a distance that is collinear with the line of charge or at a location along its perpendicular bisector.
Charged Rod Electric Field
Determine the electric field at point P, which is located a distance “a” to the left of a thin rod with a charge +Q, uniform charge density 𝜆, and length L
Notice that, if we were to place a positive point charge at point P, it would experience a force to the left from every dq or every infinitesimally small part of the wire
The direction of the electric field at point P, it will be to the left or in the negative “i" direction. Now let’s solve for the electric field
Thin Ring Electric Field
electric field caused by a uniformly charged ring of charge +Q, with radius a, at point P which is located on the axis of the ring a distance x from the center of the ring
however, all dE’s in the vertical plane cancel out because there is an equal but opposite component of dE caused by the dq on the opposite side of the ring. In the figure that is dEy.
right, if you get far enough from the uniformly charged ring, it acts like a point particle! If that sounds familiar, that is because this is true for all continuous charge distributions. If you get far enough away from them, that their own size is small by comparison to the distance, they all have electric fields which are similar to point particles.
If we use a negative charge, then the force is to the left or towards the center of the ring (SHMT). That is, the negative charge will move in simple harmonic motion about the center of the ring.
Carl Hansen's graph shows electric field along the axis of a thin ring of a uniform charge distribution, radius a = 1m, and charge Q = +1µC, with a linear region near origin which can be SHMT and inverse square law far away max electric field occurs arcounf x = 0.7m, turning point to transition bewteen the trends.
Electric Flux
Flux is defined as any effect that appears to pass or travel through a surface or substance.
Electric flux is the measure of the amount of electric field which passes through a defined area.
The equation for electric flux of a uniform electric field is:
\phi=\int EAcos\theta
ϕ is the uppercase, Greek letter phi.
E is the uniform electric field (use magnitude).
A is the area of the surface through which the uniform electric field is passing (use magnitude).
θ is the angle between the directions of E and A.
Electric flux is a scalar.
The units for electric flux are N \cdot m^2/C
determine the net electric flux of a uniform,horizontal electric field through a right triangular box and determine the electric flux through each side
When an electric field is going into a closed surface, the electric flux is negative.
When an electric field is coming out of a closed surface, the electric flux is positive.
Gauss's Law - Point Charge Electric Flux
Determine the electric flux passing through a sphere which is concentric to and surrounds a positive point charge.
*We need to use the integral equation for electric flux:
\phi=\int Eda=\frac{qencolsed}{\varepsilon0}
In other words, the electric flux through a closed Gaussian surface is:
\oint E \cdot dA = \frac{q{encl}}{\epsilon0}$$
This is Gauss’ law! Gauss’ law relates electric flux through a Gaussian surface to the charge enclosed by the Gaussian surface
A Gaussian surface is a three-dimensional closed surface
While the Gaussian surfaces we usually work with are imaginary, the Gaussian surface could actually be a real, physical surface
Typically, we choose the shapes of our Gaussian surfaces such that the electric field generated by the enclosed charge is either perpendicular or parallel to the various sides of the Gaussian surface. This greatly simplifies the surface integral because all the angles are multiples of 90° and the cosine of those angles have a value of -1, 0, or 1.
As long as the amount of charge enclosed in a Gaussian surface is constant, the total electric flux through the Gaussian surface does not depend on the size of the Gaussian surface
Gauss’ law is the first of Maxwell’s equations which are a collection of equations which fully describe electromagnetism
Notice then that, if the net charge inside a closed Gaussian surface is zero, then the net electric flux through the Gaussian surface is zero. This is why the net electric flux through the closed rectangular box in the example in our previous lesson was zero
Gauss's Law - Charged Plane Electric Field
Determine the electric field which surrounds an infinitely large, thin plane of positive charges with uniform surface charge density, σ. The electric field will be directed normal to and away from the infinite plane of positive charges.
every component of the electric field, dE, which is parallel to the plane of charges and is caused by infinitesimally small, charged pieces of the plane, dq, will cancel out,leaving only electric field components of dE which are perpendicular to the plane and directed away from the plane
We pick a Gaussian surface such that it is a cylinder with ends parallel to the plane of charges and a side parallel to the electric field and use Gauss’ law. The two ends of the Gaussian cylinder are equidistant from the charged plane
electric field is uniform and is independent of the distance from the infinite plane of charges
If you have two infinite parallel planes of charges, one with positive charge and one with negative charge, the electric field outside the planes of charges cancels out to give zero electric field outside the planes of charges. But between the two planes of charges, the electric fields add together.
In other words capacitance parallel plate capacitor is begun:)
pos charged particle moving through a constant electric field will experience an electrostatic force in the direction of the electric field, will be constant and equal to qEIn other words, the motion of a charged particle through a constant electric field will have similar characteristics to a mass moving through the constant gravitational field near the surface of a planetThis is very similar to projectile motion
Gauss's Law - Uniformly Charged Sphere Electric Field
Outside the surface of a uniformly charged sphere, the electric field is the same as if the charged sphere were a point particle
solid, uniformly charged sphere with charge Q and radius, a
Gaussian surface which is a concentric sphere with radius r > a is true of a conductor or an insulator; however, the electric field inside a conductor will be zero, and inside an insulator the electric field depends on the radius and charge distribution, and can be derived in a similar manner.
Electric Potential Difference using Integrals
The electrostatic force is a conservative force, therefore making a comparison to gravitational potential energy
we have determined the change in electric potential energy experienced by a charged particle which has moved from point A to point B in an electric field because the electrostatic force is a conservative force, this change in electric potential energy does not depend on the path taken from point A to point B
Moving a Positive Charge: vertically upward in a constant downward electric field, it will experience a positive change in electric potential energy.
Notice the negative and the dot product in the equation as we move a charge in a direction opposite the direction of the field, the direction of the displacement of the charge and the direction of the field are opposite to one another, therefore, the angle between those two directions is 180°, the cosine of 180° is negative one, which makes the change in potential energy of a positive charge positive
Electric potential defined in terms of the energy experienced by a small, positive test charge the symbol for electric potential is V
Electric potential is a scalar attribute of a vector electric field which does not depend on any electric charges which could be placed in that fieldscalar is either positive or negative for any given location; like gravitational potential energy, we need to either assign a location where V equals zero, or follow a convention for assigning the zero potential location
We will often set the initial electric potential, or electric potential at point A, equal to zeroVolt Equations
Electric Potential - Work to Move a Charge in an Electric Field in Electron Volts
A charge is moved from point A to point B via an external force changes the electric potential energy of the charge as long as there is no change in the kinetic energy of the charge, that work equals the charge of the charge multiplied by the electric potential difference the charge went through
electron volt (eV) is a unit of energy often used for very small amounts of energies, like one would use in atomic and nuclear physics (energy a charge-field system gains or losses when a charge of magnitude e (the elementary charge or the magnitude of the charge on an electron or proton) is moved through a potential difference of 1 V).
electron volt unit of energy not electric potential (volts); Also refers to a positive amount of energy, even though the electron is negative.
Parallel Plates Electric Potential Difference using an Integral
Two, large, equal magnitude charged parallel plates, the top plate has a positive charge, and the bottom plate has a negative charge . The electric field is constant in this
case and will be directed downward.If the electric potential difference between the two plates is 12 volts if the distance between the two plates is 1.0 cm defining the top plate A and the bottom plate B to determine the general equation for the electric potential difference when going from plate A to plate
a charge moving in the direction of the electric field will go through a negative potential difference and a charge moving opposite the direction of the electric field will go through a positive electric potential difference
Speed of a Proton in a Uniform Electric Field
If we have two, large, equal magnitude charged parallel plates, the top plate has a positive charge, and the bottom plate has a negative charge Then If we release a proton from the inside surface of plate A, what will the speed of the proton be right before it runs into plate B (horizontal zero line isn't needed bsubatomic particles is usually negligible)
Equipotential Surfaces
when moving at an angle relative to a uniform electric field that have the same electric potential,Points are on an equipotential surface
An equipotential surface (or line):
Has the same electric potential at every point on the surface (or line)
Is always perpendicular to the electric field - Therefore, the electric field has no component along the equipotential line - Equipotential lines are sometimes called isolines
And it takes zero work to move a charged object along an equipotential surface/lines the equipotential lines for a point charge look like this
we can use the relationship between electric potential and electric potential energy to determine the electric potential energy which surrounds and is caused by a point charge
Thin Ring Electric Field using Electric Potential
electric potential and electric potential energy are scalar values, determining those values for multiple particles uses superposition meaning that you just add all the values together
3 Properties of Conductors in Electrostatic Equilibrium
Conductors are materials where the electrons are free to move rather easily; however, when they are in electrostatic1 equilibrium, this means the charges are stationary in the object: There are four things you need to remember about conductors in electrostatic equilibrium. These are the first three:
The electric field inside a conductor in electrostatic equilibrium equals zero:
a, if the electric field inside were not equal to zero, charges would have a net electrostatic force acting on them and they would accelerate, therefore the conductor would not be in electrostatic equilibrium i. b, Notice that this means that anything inside a conductor in electrostatic equilibrium is shielded from all external electric fields -> called electrostatic shieldingAll excess charges are located on the surface (or surfaces) of the conductor:
a. Solid conducting sphere example:
i. Draw a Gaussian surface as a concentric sphere with a radius slightly smaller than the radius of the sphere
ii. Using Gauss’ law, because there is no electric field inside the conductor in electrostatic equilibrium, we know the left-hand side of the equation equals zero.
iii. Therefore, there must be zero net charge inside the Gaussian sphere and all the excess charges must be outside the Gaussian sphere.
iv. Therefore, all the excess charges are on the surface of the conductor.The electric field just outside the surface of a conductor in electrostatic equilibrium is:
a.if the electric field had a component parallel to the surface of the conductor, the charges would move, and the conductor would no longer be in electrostatic equilibrium. Therefore, the electric field at the surface of a conductor in electrostatic equilibrium must be perpendicular to the surface
i.Because equipotential surfaces are always perpendicular to the electric field, the surface of a conductor in electrostatic equilibrium must be an equipotential surface.
b. If we zoom way in on the surface of the conductor in electrostatic equilibrium, we can draw a Gaussian cylinder with its cylindrical axis normal to the surface of the conductor.The fourth thing you need to remember about conductors in electrostatic equilibrium is:
• "Irregularly Shaped Conductors in Electrostatic Equilibrium"