Chapter 17

Net electric charge – when an object gains an electric charge, allowing them to attract other
objects.
o Objects with equal charge repel each other. Objects with a certain charge only attract
objects with opposing charges (i.e Glass and plastic ( negative and positive). All charges
regardless are equal in magnitude.
o Magnitude of charge: e = 1.602176487(40) x 10-19 C (one proton)
 Negative of this is one electron
 All electric charge is a multiple of this (quantized)
o Principle of charge convservation: The algebraic sum of all electric charges in any closed
system is constant.
 If a charge is transferred to another object, it’s divided equally (i.e. ball
question).
o Reviewing in counting atom numbers:
 The number of neutrons equal the number of neutrons.
 The net charge is the # of protons minus # of neutrons. Negative charge (-) and
positive charge (+). Same value is neutral.

Ion – atom that has lost or gained at least one electron.
o Positive ion – ion with a positive charge after losing electrons
o Negative ion – ion with a negative charge after losing electrons

Conductor - material that conducts and transfers electric charge
o Most metals
Insulator – material that stops the transfer of electric charge
o Most nonmetals
o All the electrons in insulators are bound tightly, so there’s no free electrons to attract
stuff (such as with conductors)
Semiconductor – intermediates between the two
o Silicon (can be manipulated to control its conductivity)
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Electric charges move from positive to negative (in liquids and gases) (solid loses electron (X+;
positive charge/electron loss) and liquid/gas gets electron (X-; negative charge/electron gain))
Neutral objects don’t attract anything. They NEED a charge.
Induction - Charging an object without sacrificing any of its charge. (There’s no direct transfer,
they’re only in the same vicinity)
o When two objects of identical charges interact without touching, the shared charge
moves and the opposing/free charge moves in its place to repel them (induced charges).

o

The induced charges are small, and offers a weak attraction between the two
objections, and not a full repel.
For the affected object, it needs another medium to rid of its excess charge to balance it
out. By doing this, it maintains the same exact behavior when interacting with other
objects (i.e always attracts/repels, extra charge is in the ground)
Polarization –
Coulomb’s Law – The magnitude of force (F) between two charges/Coulomb’s (q1,q2) at r distance is
proportional to the product of both charges, and inversely proportion to the inverse of the square
distance between them (1/r2).
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Acts in accordance to Newton’s third law: If object (charge 1) acts a force on another (charge 2),
the receiving object exerts a force of equal magnitude and opposite direction back on the first
object.
o The direction of the force is the same as the line joining them.
o The higher the mass of the object, the slower the acceleration.
o Remember: the balls experience the same magnitude of force. It having a higher integer
of charge (+q) doesn’t matter.
Same with regular charges: opposite attract, identical repel.
|𝑞1 𝑞2 |
)
𝑟2
F=k(
k=
1
4Π𝐸0
= 8.988 x 109 N . m2/C2 (1 Coulumb)
E0 = 8.854 x 10-12 C2/N . m2 (electric constant)
Principle of superposition – When two charges exert simultaneous force on a third charge, the total
force acting on that is the vector sum of the forces both charges exert individually.
How to solve a Coulomb’s Law problem:
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
Convert everything into coloumbs beforehand. DON’T FORGET! (1 microC = 10-6 C)
Calculate the vector sum of two acting forces on a third charge (review vector sum addition)
When calculating a two charge system, calculate both opposing charges, then subtract the smaller
charge from the bigger charge.
[spot for vector addition tutorial]
[try example 3]
Electric field – An electric charge in the space around a charged object.
Test charging – A method to mark the presence of an electric field.

When a charged object’s former position (P) is acted on by an electric force (𝐹⃗ ′) generated by
another charged object the former interacted with. The action creates a test charge in P’s place
(𝑞′). (It can either be positive or negative).
o REMEMBER, THE RESPONSIVE FORCE/ACCELERATION SHOT AT (P) IS OPPOSITE TO THE
ELECTRIC FIELD’S CHARGE AND EQUAL TO ITS ACCELERATION.
o A force on the positive test charge points in the direction of the electric field (outward).
A force on the negative test charge points in the opposite direction (inward).
o In an electrostatic situation (if an electric field is within a conductor) the charges do not
move, making the electric field inside it zero.
⃗
𝐹′
𝐸⃗⃗ = (electric field (N/C))
𝑞′
Calculating electric fields/electric field due to a point charge

When you want to find the magnitude (E) of an electric field at point (P) due to point charge (q)
generated from a source point (S), at a certain distance (r) from the test point (P):
1
𝑞
𝐸⃗⃗ = (
) ( 2 ) 𝑁/𝐶
4𝜋𝐸0 𝑟
*first equation is k. See above.
Draw out your problems before solving them
1. Use the above equation to find E1
2. Graph the intermediate point, and find the E1 and E2.
3. Get the vector sum of both charge points.
a. Subtract the smaller charge from the bigger charge.
b. Remember: Pos charge out (E will always be positive). Negative charge in (E will always
be negative). DO NOT CHANGE THE SIGNS.
For a three charge system with electric field point:

Electric field line – imaginary line in a space that curves so that each is tangent to the direction of the
electric field vector of that point (AKA lines of force).

Electric flux: The amount of electric field lines that flow through an Area.
#2: The reason there’s no electric field on the repel, is cause the identical charges cancel each other out.
There’s no imbalance, so it’s just empty space. Meaning there’s no electric fields in the graph. There’s
no opposing charge to draw it in.
Gauss’s Law – law of electric flux

Electric flux – The amount of electric field lines flowing through an Area.
o Electric fields always flows traight ahead.
Perpendicular
Tilted Perpendicular
Parallel (90o tilt)
Φ𝐸 = 𝐸𝐴
Φ𝐸 = 𝐸𝐴𝑐𝑜𝑠∅
Φ𝐸 = 𝐸𝐴𝑐𝑜𝑠90𝑜 = 0
Standard electric flux equation: Φ𝐸 = 𝐸𝐴𝑐𝑜𝑠∅ 𝑁 ⋅ 𝑚2 /𝑐
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E = field; A = Area
Just use this one (cos0 = 1; a parallel surface is 0 to begin with)
How to solve disk problem on presentation:
1. Fine the area based on the radius (pi(r2)), then plug it into the base problem. (Find the total flux
of the disk)
2. Apply the same, though since its parrel the cos is relegated of 0, so its 0 (Find the total flux if
plane itself is parallel)
3. Apply the same equation, though the cos is 1 do to the axis being parallel (Find the total flux if
the axis itself is parallel)
Charge in the box
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Flux in a box acts similar to an electric field and acts as long as the box has an electric charge.
o If the box is positively charged, the field lines spread outward (positive value)
o If the box is negatively charged, the lines spread inward. (negative value)
If there is no charge or not, the flux is zero, even if there is a stated charged in the box.
How to solve a flux box (sphere) problem
1. Use Φ𝐸 = 𝐸𝐴, E = electric field equation, A = Surface area of a sphere
2. Due to the nature of the problem, you can cancel a grand majority of the problem as shown
below:
Gauss’s Law – the total electric flux through a closed surface is proportional to the net charge inside the
(Gaussian) surface.
𝜙𝐸 = 4𝜋𝑘𝑄𝑒𝑛𝑐𝑙
*Remember what k is (scroll above)
2nd example:
3.0 microC, distance – 0.2 m, 0.5 m – distance from point charge
1. Find the radius (Gaussian surface radius would be the distance (0.2 m)
2. Set up the
3. Plug it into the Flux problem