Quiz 1 - Chapter 18 Studying Flashcards
Four identical metallic objects carry the following charges: +1.6, +6.2, −4.8, and −9.4 μC. The objects are brought simultaneously into contact, so that each touches the others. Then they are separated. (a) What is the final charge on each object?
When identical metal objects are brought into contact, the charge is distributed evenly between them.
Hence you add up all these charges and divide by 4 to find the final charge on each object (which is the same at -1.6 microcolumbs)
Two very small spheres are initially neutral and separated by a distance of 0.50 m. Suppose that 3.0 × 1013 electrons are removed from one sphere and placed on the other. (a) What is the magnitude of the electrostatic force that acts on each sphere?
Answer lies in knowing that the only charges that matter are those that are being transferred, and after the transfer the spheres have equal magnitude of charges (q1 = q2) but opposite signs
1) Calculate q=ne
You should get 4.8x10^-6
2) Calculate the force using colombs law, you should get 0.83N
Suppose a single electron orbits about a nucleus containing two protons (+2e), as would be the case for a helium atom from which one of the two naturally occurring electrons is removed. The radius of the orbit is 2.65 × 10−11 m. Determine the magnitude of the electron’s centripetal acceleration.
You use colombs law and then the F = ma to find the acceleration
But first to find the force
q1 = charge of electron which is simply e
q2 = charge of proton which is 2e
How do you find the magnitude of the net electric field?
Enet = unfirom electric field + electric field created by point charge
A uniform electric field exists everywhere in the x, y plane. This electric field has a magnitude of 4500 N/C and is directed in the positive x direction. A point charge −8.0 × 10−9 C is placed at the origin. Determine the magnitude of the net electric field at (a) x = −0.15 m, (b) x = +0.15 m, and (c) y = +0.15 m
In each of these examples calculate the electric field due the point charge which is 3196N becuase the r is the same for all three examples
In general, electric field points from positive to negative so this will be in the right direction (as also stated). And it points towards the charge
a)
At -0.15, one vector left and one right so substractive(1304.44)
b) At -0.15, both vectors are pointing right so they are addative (7695.56)
c) There is the netx and nety, so you use pythagorean theorem to find the net
Enet = sqrt(4500^2)+(3104^2) = 5500
You should note that the uniform electric field has no y direction and the field caused by the point charge has to x direction.
so we have Euniform (in +x) and Eq )in -y) hence why we have to do trig
If the force points upward, and the charge is negative, what direction is the electric field
Negative and thus downwards according to the equation E = Fq
What do you KEEP forgetting with the electric field equation?
PLEASE SQUARE THE RADIUS
A charge +q is located at the origni while an identical charge is located on the x axis at x = 0.5m. A third charge of +2q is located on the x axis at such a place that the net electrostatic force on the charge at the origin doubles, its direction remaining unchanged. where should the third charge be located?
The main idea is that the third charge should increase total force on origin without changing direction, so must act same direction as force from charge at x = 0.5m (F1)
SUe coulombs law to calcualte F1 and F2
F1 = kq^2/0.25
F2 = k2q^2/x^2
Net force is
F1 + F2 = 2F1
F1 = F2
k2q^2/x^2 = kq^2/0.25
2/x^2 = 1/0.25
x = 0.71
In an experiment in space, one proton is held fixed and another proton is released from rest a distance of
2.50 mm away. (b) Sketch qualitative
(no numbers!) acceleration–time and velocity–time graphs of the released proton’s motion
a) The particle is always accelerating because of the repulsion, but the rate at which it accelerates decreases over time since coulomb’s force weakens with distance
Since the particle is always acceleration that means its speed is always increasing
As long as acceleration is positive speed is increasing. Proton always being pushed away so it speed keeps increasing but at a decreasing rate.
Acceleration looks liek decreasing exponential and speed is like root x function levelling off
Bella is awesome <3
What is an eaiser equation for electric flux?
Φ E=EA
Flux = electric field x area
EA = Q/E0
two charges, -Q and -3Q are distance l apart. third charge q in between them. what must be its placement and magnitude so that the first two charges are in equilibrium
1) Find the equations for the left charge (set them equal to each other because equilibrium means repulsive and attractive cancel out)
kQq/x^2 = k3Q^2/l^2
2) Find the equations for the right charge
k3Qq/(l-x)^2 = k3Q^2/l^2
3) Isolate for x from equation 1 (this gives you 0.37l which is the distance) and sub into equation 2 (this gives you 0.40Q which is the mag of the charge)
Electron accelerated from rest between two parallel charged plate, in uniform electric field of E = 1.45 x 10^4. The plates are separated by 0.016m. What speed does electron leave hole in plates?
You need the kinematic equation
v^2 = v0 + 2ax
To solve for acceleration
F = ma
F = qE
a = qE/m
In the Bohr model of the hydrogen atom, the electron is in orbit about the
nuclear proton at a radius of 5.29x10 ̶11m.
Determine the speed of the electron, assuming the orbit to be circular
1) Use F = kq1q2/r^2 to get 8.22 x 10^-8
2) Use F = mv^2/r (cenripetal acceleration because circular orbit)
What does F12 mean?
Force on q1 due to Rq2
When to distinguish between using
E = k|q|/r^2
E = F/q
The first one is for field created by point charge
Point charge: charge used to create electric field, like electron or proton
The second one is for firled experienced by test charge
Test charge: charge used to measure electric field, ideally infintestimally small
EXAM PROBLEM
Imagine a square ABCD. There is a point charge +q in the center. Find the magnitude of this charge at B in terms of q so that the Enet = 0 at D
So at B has to be negative charge to counteract this
Now you calculate the electric field of +q on D and then B on D
E = kq/r^2 and E = kqb/2r^2
The first one is from B on D and second one from center on D
Use pythagoran theorem to get r = sqrt(r^2+r^2)
then you end up with sqrt2(r)
you square this in the equation to get 2r^2
The radius of the diagonal of a square can be found using pythagoran theorem. then you didvide by 2 to get just the distance between +q and D and you get simply r
Solve and you find that qB = 2q
EXAM PROBLEM
In the rectangle in the drawing, a charge is to be placed at the empty corner to make the net force on the
charge at corner A point along the vertical direction. What charge (magnitude and algebraic sign) must be
placed at the empty corner?
btw the rectangle looks like 3 point charges of +3uc and the charge in question is at the top right. the sides are d and 4d respectively
FAU must point up to the right in order to counteract qA which is pointing to the right in order for the net force along the horizontal to be 0
Thus
FaU x component and Fa2 are equal to each other
Fau = kqAqU/(4d)^2 _ d^2
FA2 = kqAq2/4d^2
The horizontal component of the force is costheta , eliminating the similar variables you end up with
qU/17cosx = q2/16
cos x = 4/sqrt17
the charge ends up being 3.3x10^-6
EXAM PROBLEM
two charges q1 (-) and q2 (+) produce electric field at point P. the distance from each of the charges to P is 0.7m such that an isosceles triangle is produced. the angle from p relative to the horizontal to each of the charges is 30 degrees. E1 points from P towards q1, E2 points away from P the other side. net electric field points upwards. find the net electric field
Electric field points from positive to negative which explains directions of E1 and E2
E1x = E2x because horizontal components cancel because they point in opposite directions (so don’t even consider cause Enet just points in the y direction
Enety = E1y + E2y
which turns out to be
E net = 2E1sin30
E1 = kq/r^2. If you assume that q = e then you can figure out the E net
EXAM PROBLEM
A small spherical insulator of mass 8.00x10–2 kg and charge
+0.600 µC is hung by a thin wire of negligible mass.
A charge of – 0.900 µC is held 0.150 m away from the
sphere and directly to the right of it, so the wire makes an
angle θ with the vertical (see the drawing). Find (a) the angle
θ and (b) the tension in the wire
a)
Label the forces in the x and y noting that the electrostatic force acts to the right and the weight acts downward, the tension acts an angle upwards to the left
Note that Tx = Tsintheta cause the angle is in a weird position
Tsinx = kq^2/r^2
Tcosx = mg
Divide both equations to cancel out T (subsitution also works but this is a lot more efficient)
tanx = kq^2/mgr^2
solve for x by subbing in all known values
tan-1[(8.99x10^9)(0.6 x10^-6)(0.9x^10-6)/8.00x10^-2(9.80)(0.150)^2)
= 15.4 degrees
b) To find the tension, you can simply sub into
T = mg/cosx
= 8.00x^-2*9.80/cos15.4
0.813N
EXAM PROBLEM
The drawing shows an electron entering the lower left side of
a parallel plate capacitor and exiting at the upper right side.
The initial speed of the electron is 7.0x106 m/s. The capacitor
is 2.00 cm long, and its plates are separated by 0.150 cm.
Assume that the electric field E between the plates is uniform
everywhere and find its magnitude
Also how would this problem change if the particle entered midway
When the electric field is uniform, the acceleration is constant and hence kinematic equations can be used.
Electric field acts in vertical direction so horizontal motion is unaffected.
a) so use t = x/v0 (no horizontal acceleration electric field acts vertically downward
t = 0.02/7.0x10^6
= 2.86 x 10^-0
b) F = ma
a = F/m
a = qE/m
c) y = 1/2at^2 (bc we assume the inital velocity in the y direction is 0)
y = 1/2(qE/m)t^2
This vertical displacement is equal to the separation of the plates
s = 1/2(qE/m)t^2
d) now isolate for E
E = 2ms/qt^2
and subsitute known value. you should get the answer to be 2088N/C
Followup: If it entered midway then you take the s vaue dividedby 2
