Chapter 18 - Electric Forces & Fields Flashcards
(20 cards)
The isolated point charge of
q = +15μC is in a vacuum. The
test charge is 0.20 m to the right
and has a charge qo
= +0.8μC.
Determine the electric field at
point P.
F = kqq/r^2
Use q = 0.8 and q = 15
= 2.7N
Step 2
Find electric field using test charge
E = F/q0
= 3.4 x10^6
ORRRR in one step you can use the equation for poitn charges
E = kq/r^2
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)
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
Two positive point charges, q1
= +16 μC and q2
= +4.0 μC are separated
in a vacuum by a distance of 3.0 m. Find the spot on the line between
the charges where the net electric field is zero
E1 = E2
E = kq/r^2
16k/d^2 = 4k/(3-d)^2
Solve and isolate for d to get 2m
A positive charge is suspended at
the center of
a hollow,
electrically neutral,
spherical conductor
.
Show that this charge
induces
(a)
a charge of
–
q on the
interior surface and
(b)
a charge of
+
q on the
exterior surface of
the conductor
.
a)
Electric field inside conductor must be zero, to cancel field from central +q, inner surface must develop a -q charge.
b) Total conductor must remain neutral so if -q is induced inside, conductor must develop +q on outer surface
Square ABCD with +q in the center
a) At which corner is a second
charge located to fulfill the
condition: Enet= 0 at
D
b. Is the second charge
positive or negative
c. Does the second charge
have a greater, equal, or
smaller magnitude as
q
d. Find the magnitude of this
charge in terms of
q
a) Opposite across from D so that would be B
Where it can oppose the field from q at D
b) To cancel the charge, the second charge must be negative
c) Since the second charge is further away, its field is weaker. To balance the closer field, it must have a greater magnitude
d) Assume the side is a
diagonal is a^2 + a^2 = 2a2
Take sqrt to find diagonal to be √2a
for q
E = kq/r^2 and E
An alternative way to think about it:
Using E = kq/r^2, the difference is teh ratio of the r squared
So to +q is 1 and to B is 2 so (1/2)^2 ratio which means B has to be 4 times the charge. Shows the charge has to be 4x stronger
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
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
A charge +q is located at the origin, while an identical charge is located on the x axis at x=+0.50 m. 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?
Force at q0 is leftwards, so to maintain this charge, you have to add the third charge to the right of both charges, to the right of x = 0.5m
Frce due to q+ charge at x = 0.5 + force due to 2q+ charge at x = +d = twice force due to +q charge at x = 0.5m
kqq/(0.5)^2 + k2qq/d^2 = 2kqq/(0.5)^2
Solving and isolating d^2 = 2(0.5)^2
d = 0.71
When given Fnetx and Fnety, how do you find Fnet and the direction?
Fnet = sqrt(Fx^2 + Fy^2)
Pythagorean Theorem
tanx = Fy/Fx
In parallel plate capacitor what is the direction of acceleration?
Parallel to field lines
What is a gaussian surface?
Imaginary surface with radius r, must be closed and usually is spherical
What direction is the electric field?
Positive to negative
High to low
How do you convert to electron volts?
Divide by 1.6 x 10^-19
What is the energy when electron accelerated from rest through potential difference
eV or qV
EXAM QUESTION
two point charges q1 and q2 are held 6cm apart. an electron released at point A that is equidistant from both charges at distance c = 4cm undergoes an inital acceleration of 7.60x10^17 directly downward, parallel to the line connecting q1 and q2
c) calculate the magnitude of q1 and q2
So the force from q1 on e points down and to the left and from q2, down and to the right. Thus the horizontal components cancel and we are left with 2F directly downwards
Nety = 2Fcosx (cosx becuase it’s a weird thingie now)
ma = 2(kqe/r^2)cosx
9.117.60(0.050.05)
/
28.991.60cos53.1
solving for x
you get tanx = 0.04/0.03
x = 53.1
A negative point charge lies along figure. Find magnitude and direction of electric field at point P, which lies 6cm from -2uc charge measured perpendicular to line connecting 3 charges.
So lines are oriented towards negative charge (field points from positive to negative)
Calculate teh electric field of E1 and E3 which are equal
E = kq/r^2
E1 = E3 = 4.49 x 10^6
E2 = 4.99 x 10^6
Net y = E1y = E3y
Netx = 2E1cos53.1 + E2
= 1.04 x 10^7
