concept questions with answers coulomb s law,...
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Concept Questions with Answers
Coulomb’s Law, Electric Field, Discrete Charge Distributions
8.02
W01D2
http://www.nytimes.com/video/science/100000003489464/out-there-raining-fire.html?playlistId=100000002906445
8.02 W01D2
CQ: 5 Equal Charges
Six equal positive charges q sit at the vertices of a regular hexagon with sides of length R. We remove the bottom charge. The electric field at the center of the hexagon (point P) is:
1.E =
2keqR2 j
2.E = −
2keqR2 j 5.
E = 0
4.E = −
keqR2 j
3.E =
keqR2 j
CQ Ans.: 5 Equal Charges
Answer 4. Electric fields of the side pairs add to zero (symmetry). Electric field at center is only due to top charge, which is a distance R away. Electric field at P points downward.
Alternatively: “Added negative charge” at bottom R away, pulls field down
E(P) = −
keqR2 j
CQ: Electric Field
1. Between the two charged objects. 2. To the right of the charged object on the right. 3. To the left of the charged object on the left. 4. The electric field is nowhere zero. 5. Not enough info – need to know which is positive.
Two charged objects are placed on a line as shown below. The magnitude of the negative charge on the right is greater than the magnitude of the positive charge on the left, . Other than at infinity, where is the electric field zero?
qL > 0 qR < 0
qR > qL+
qR > qL
CQ Ans.: Electric Field Answer: 3. To the left of the positively charged object on the left
Between: field points from positively charged object to negatively charged object On right: field is dominated by negatively charged object On left: electric field due to close smaller positively charged object will cancel electric field due to further larger negatively charged object.
http://public.mitx.mit.edu/gwt-teal/PCharges.html
CQ: Field Lines
1. Directions of forces that exist in space at all times.
2. Directions in which positive charges on those lines will accelerate.
3. Paths that charges will follow. 4. More than one of the above.
Electric field lines show:
CQ Ans.: Field Lines
Answer: 2. Directions positive charges accelerate. Note: This is different than flow lines (3). Particles do not move along field lines. (1) is not correct because we don’t know sign of charge so we don’t know direction of force
Concept Questions with Answers
8.02
W02D1
Electric Dipoles and Continuous Charge
Distributions
W02D1
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CQ: Field Lines
1. Directions of forces that exist in space at all times.
2. Directions in which positive charges on those lines will accelerate.
3. Paths that charges will follow.
4. More than one of the above.
Electric field lines show:
CQ Ans.: Field Lines
Answer: 2. Directions that positive charges accelerate. Note: This is different than flow lines (3). Particles do not move along field lines. (1) is not correct because we don’t know sign of charge so we don’t know direction of force
CQ: Dipole in Non-Uniform Field
An electric dipole sits in a non-uniform electric field.
Due to the electric field this dipole will feel: 1. force but no torque 2. no force but a torque 3. both a force and a torque 4. neither a force nor a torque
13
+
p
q+
q
E
CQ Ans.: Dipole in Non-Uniform Field
Answer: 3. both force and torque
14
+
p
q+
q
E
Because the electric field is non-uniform, the forces on the two point charges do not cancel. The dipole wants to rotate to align with the field – there is a torque on the dipole as well
A uniformly charged ring of radius a has total charge Q. Which of the following expressions best describes the electric field at the point P located at the center of the ring?
CQ: Electric Field of a Uniformly Charged Ring
1.!E(P) = −ke
λadθa3
θ=0
θ=2π
∫ i
2.!E(P) = ke
λadθa3
θ=0
θ=2π
∫ i
3.!E(P) = −ke
Qa2 i
4.!E(P) = +ke
Qa2 i
5.!E(P) =
!0
.P
Qi
y
z
j
a
x
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CQ Ans.: Electric Field of a Uniformly Charged Ring Answer
Answer: By symmetry, the electric field from all infinitesimal segments of the ring add to zero at the center of the ring, (vector addition).
5.E(P) =
0
16
.P
Qi
y
z
j
a
x
Concept Questions with Answers
8.02
W02D2
Gauss’s Law
8.02 W02D2
CQ: Flux thru Sphere
1. positive (net outward flux). 2. negative (net inward flux). 3. zero. 4. not well defined.
The total electric flux through the spherical surface shown in the figure below is
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spherical surface
+Q
CQ Ans.: Flux thru Sphere
We know this from Gauss’s Law: No enclosed charge à no net flux. Flux in on left cancelled by flux out on right
Answer: 3. The total flux is zero
ΦE =E ⋅dA
closedsurface S
∫∫ =qenclosed
ε0
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spherical surface
+Q
CQ: Sign of Flux
1. positive. 2. negative. 3. zero. 4. Not well defined.
The electric flux through the planar surface below (positive unit normal to left) is:
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+Q Q
n
CQ Ans.: Sign of Flux
The field lines go from left to right, opposite the assigned normal direction. Hence the flux is negative.
Answer: 2. The flux is negative.
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+Q Q
n E = E+ +E
CQ: Gauss’s Law The grass seeds figure shows the electric field of three charges with charges +1, +1, and -1, The Gaussian surface in the figure is a sphere containing two of the charges. The electric flux through the spherical Gaussian surface is 1. Positive 2. Negative 3. Zero 4. Impossible to determine without more information.
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CQ Ans.: Gauss’s Law Answer 3: Zero. The field lines around the two charged objects inside the Gaussian surface are the field lines associated with a dipole, so the charge enclosed in the Gaussian surface is zero. Therefore the electric flux on the surface is zero. Note that the electric field E is clearly NOT zero on the surface of the sphere. It is only the INTEGRAL over the spherical surface of E dotted into dA that is zero.
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CQ: Spherical Shell
1. Zero 2. Uniform but Non-Zero 3. Still grows linearly 4. Some other functional form (use Gauss’ Law) 5. Can’t determine with Gauss Law
We just saw that in a solid sphere of charge the electric field grows linearly with distance. Inside the charged spherical shell at right (r<a) which expression best describes the magnitude of the electric field?
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+Q
uniform charged spherical shell
a
CQ Ans.: Flux thru Sphere
Spherical symmetry Use Gauss’ Law with spherical surface. Any surface inside shell contains no charge therefore no flux. By spherical symmetry: electric field must be zero.
Answer: 1. Zero
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+Q
uniform charged spherical shell
a
CQ: Superposition
Three infinite sheets of charge are shown above. The sheet in the middle is negatively charged with charge per unit area -2σ, and the other two sheets are positively charged with charge per unit area +σ. Which set of arrows (and zeros) best describes the electric field?
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CQ Ans.: Superposition
Answer 2 . The fields of each of the plates are shown in the different regions along with their sum.
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Concept Questions with Answers
8.02 W03D2
Work, Potential Energy and Electric Potential
W03D2
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CQ: Motion of Positive Charged Object in External Electric Field
1. from higher to lower electric potential resulting in an increase in potential energy.
2. from higher to lower electric potential resulting in a decrease in potential energy.
3. from lower to higher electric potential resulting in an increase in potential energy.
4. from lower to higher electric potential resulting in a decrease in potential energy.
If a positively charged particle is released from rest in an electric field, the charge will move
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+++++++
E
+
CQ Ans.: Motion of Positive Charged Object in External Electric Field
Answer: 2. When positively charged particle is released from rest in an electric field, it will always move in the direction that decreases the potential energy of the system. Because q > 0 and ΔU < 0, it must move from higher to lower electric potential,
ΔU = qΔV < 0⇒ΔV < 0.
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CQ: Motion of Negative Charged Object in External Electric Field
1. from higher to lower electric potential resulting in an increase in potential energy.
2. from higher to lower electric potential resulting in a decrease in potential energy.
3. from lower to higher electric potential resulting in an increase in potential energy.
4. from lower to higher electric potential resulting in a decrease in potential energy.
If a negatively charged particle is released from rest in an electric field, the charge will move
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+++++++
E
CQ Ans.: Motion of Charged Objects
Answer: 4. When a negatively charged particle is released from rest in an electric field, it will always move in the direction that decreases the potential energy of the system. Because q < 0 and ΔU < 0, it must move from lower to higher electric potential,
ΔU = qΔV < 0⇒ΔV > 0.
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CQ: E from V
1. Ex > 0 is positive and Ex < 0 is positive 2. Ex > 0 is positive and Ex < 0 is negative 3. Ex > 0 is negative and Ex < 0 is negative 4. Ex > 0 is negative and Ex < 0 is positive
The above shows potential V(x). Which is true?
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CQ Ans.: E from V
Answer: 2. Ex > 0 is positive and Ex < 0 is negative. E is the negative slope of the potential, positive on the right, negative on the left,
Translation: “Downhill” is to the left on the left and to the right on the right.
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Concept Questions with Answers
8.02
W04D1
Electric Potential and Gauss’ Law Equipotential Lines
Concept Questions
W04D1
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CQ: Potential Difference
Consider two oppositely uniformly charged parallel plates. The electric potential difference V(B) – V (A) between the points A and B that lie outside the plates is 1. Positive. 2. Negative 3. Zero.
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CQ Ans. : Potential Difference
Answer 2. Electric field points from higher potential to lower potential . So V(B) – V (A) < 0.
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Consider the point-like charged objects arranged in the figure below. The electric potential difference between the point P and infinity and is From that can you determine the functional form for the electric field at the point P?
CQ : E from V
V (P) = − kQ a .
1. Yes, its kQ/a2 (up) 2. Yes, its kQ/a2 (down) 3. Yes in theory, but I don’t know how to take a gradient 4. No, you can’t get the electric field at point P from just
knowledge of the electric potential at point P 41
P
Q Q +Q
y+
x+a
a a
CQ Ans.: E from V
Answer 4. No, you can’t get the electric field at point P from just knowledge of the electric potential at point P. The electric field is the gradient (spatial derivative) of the potential. Knowing the potential at a single point tells you nothing about its derivative. People commonly make the mistake of trying to do this. Don’t!
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CQ: Equipotentials We show the grass seeds representation of the electric potential of two point charges. Which of the following statement(s) is correct?
1. Both charges have the same sign and equal magnitude. 2. Both charges have the same sign and unequal magnitude. 3. The charges have opposite signs and equal magnitude. 4. The charges have opposite signs and unequal magnitude.
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CQ Ans.: Equipotentials
Answer 2. The equipotential lines near point-like charged objects are circles. The charged objects have the same sign of charge with unequal magnitudes. If the charges were opposite signs then there would be a zero point between them, and the equipotential corresponding to the zero point would extend in a perpendicular direction to a line passing through the two charged objects. If the charged objects had equal magnitude then the equipotential lines would be symmetric about an axis passing perpendicular to a line passing through the two charged objects.
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CQ: Two Point Charges
1. is positive. 2. is negative. 3. is zero. 4. can not be determined – not enough info is given.
The work done by the electrostatic interaction when a positively charged particle that starts from rest at infinity and is moved to the point P midway between two fixed charges of magnitude +Q and –Q
+
+
d / 2 d / 2
infinity
+Q Q
q > 0
P
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CQ Ans.: Two Point Charges
Answer 3. Electrostatic work done in moving the charged particle from ∞ to P is because the potential is zero at both ∞ and the point P. The potential at P is zero because equal and opposite potentials are superimposed from the two point charges (remember: V is a scalar, not a vector).
W = −ΔU = −qΔV = 0
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+
+
d / 2 d / 2
infinity
+Q Q
q > 0
P
Concept Questions with Answers
8.02 W04D2
Conductors and Insulators Capacitance & Capacitors
Energy Stored in Capacitors
W04D2
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CQ: Point Charge in Conductor
1. is initially uniform and does not change when the charge is moved.
2. is initially uniform but does become non-uniform when the charge is moved.
3. is initially non-uniform but does not change when the charge is moved.
4. is initially non-uniform but does change when the charge is moved.
A point charge +Q is placed inside a neutral, hollow, spherical conductor. As the charge is moved around inside, the surface charge density on the outside
spherical conducting shell
+ Q
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CQ Ans.: Q in Conductor
E = 0 in conductor à non-uniform charge density
on inside
Answer 1. charge density is initially uniform and does not change when the charge inside is moved.
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Q+ +
outer uniform charge density
inner non-uniform negative charge density
out
in
E = 0 in conductor à No “communication” between inside charges and therefore charge density on outside remains uniformly distributed
CQ: Point Charge in Conductor
1. is zero and does not change 2. is non-zero but does not change 3. is zero when centered but changes 4. is non-zero and changes
A point charge +Q is placed inside a neutral, hollow, spherical conductor. As the charge is moved around inside, the electric field outside
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spherical conducting shell
+ Q
CQ Ans.: Q in Conductor
E = 0 in conductor à -Q on inner surface Charge conserved à +Q on outer surface
Answer: 2. is non-zero but does not change.
Q+ +
Q
+Q
inner surface
outer surface
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E = 0 in conductor à No “communication” between –Q & +Q à + Q remains uniformly distributed on outer surface so E outside of shell stays unchanged
CQ: Point Charge in Conductor
1. Q. 2. Q + q. 3. q. 4. Q - q. 5. Zero.
A point charge +q is placed inside a hollow cavity of a conductor that carries a net charge +Q. What is the total charge on the outer surface of the conductor?
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CQ Ans.: Point Charge in Conductor
Answer 2. Choose Gaussian surface inside conductor. Electric field is zero on Gaussian surface so flux is zero. Therefore charged enclosed is zero. So an induced charge –q appears on cavity surface. Hence an additional charge of +q appears on outer surface giving a total charge of Q + q on outer surface. 54
CQ: Metal Spheres Connected by a Wire
Two conducting spheres 1 and 2 with radii r1 and r2 are connected by a very long thin wire. What is the ratio of the charges q1/q2 on the surfaces of the spheres? You may assume that the spheres are very far apart so that the charge distributions on the spheres are uniform. 1. (r1 / r2)2 2. (r2 / r1)2 3. r1 / r2 4. r2 / r1
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CQ Ans.: Metal Spheres Connected by a Wire
Answer 3. The wire and two spheres are all at the same potential. With V=0 at infinity, the potential on the surface of each sphere is the same as if the charge on each sphere were located at the origin. Set the potentials equal V1 = kq1 / r1 = V2 = kq2 / r2. Therefore q1 /q2= r1 / r2.
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CQ: Changing Dimensions
A parallel-plate capacitor is charged until the plates have equal and opposite charges ±Q, separated by a distance d, and then disconnected from the charging source (battery). The plates are pulled apart to a distance D > d. What happens to the magnitude of the charge on the plates? 1. Q increases. 2. Q is the same. 3. Q decreases.
!
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CQ Ans.: Changing Dimensions
Answer: 2. With no battery connected to the plates the charge on them has no possibility of changing.
!
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CQ: Changing Dimensions
A parallel-plate capacitor is charged until the plates have equal and opposite charges ±Q, separated by a distance d, and then disconnected from the charging source (battery). The plates are pulled apart to a distance D > d. What happens to the magnitude of the potential difference V? 1. V increases. 2. V is the same. 3. V decreases.
!
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CQ Ans.: Changing Dimensions
Answer: 1. With no battery connected to the plates the charge on them has no possibility of changing. In this situation, the electric field does not change when you change the distance between the plates, so: V = E d. As d increases, V increases.
!
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Concept Questions with Answers
8.02 W05D1
Dielectrics and Conductors as
Shields
W05D1
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CQ: Dielectric Capacitor Charge A parallel plate capacitor is charged to a total charge Q and the battery is disconnected. A slab of material with dielectric constant κ is inserted between the plates. The charge stored in the capacitor
1. increases. 2. decreases. 3. stays the same.
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+ + + ++++++Q
Q
CQ Ans.: Dielectric Capacitor Charge
Because the capacitor is disconnected from the battery, there is no conducting path for the charge to leave either plate.
Answer: 3. The charges ±Q on the plates stays the same
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+ + + +++++
+ +++
+Q
Q
Qind
Qind+
CQ: Dielectric Capacitor Energy I
1. increases. 2. decreases. 3. stays the same.
A parallel plate capacitor is charged to a total charge Q and the battery is disconnected. A slab of material with dielectric constant κ is inserted between the plates. The energy stored in the capacitor
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+ + + ++++++Q
Q
CQ Ans. Dielectric Capacitor Energy I
The dielectric reduces the electric field and hence reduces the amount of energy stored in the field. The easiest way to think about this is that the capacitance is increased while the charge remains the same so Also from energy density:
Answer: 2. Energy stored decreases
uE ,0 =
12ε0E2 ⇒ 1
2κε0( ) E
κ⎛⎝⎜
⎞⎠⎟
2
= 12ε0E2
κ< uE ,0
U = Q2 / 2C
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CQ: Dielectric Capacitor Energy II A parallel plate capacitor is charged to a total charge Q. While the battery is still connected a slab of material with dielectric constant κ is inserted between the plates. The energy stored in the capacitor
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1. increases. 2. decreases. 3. stays the same.
+ + + ++++++Q
Q
Vbatt
CQ Ans.: Dielectric Capacitor Energy II Answer: 1. The energy stored before the dielectric is inserted is After the dielectric is inserted, the potential difference between the plates remains the same because the plates are connected to the battery. The capacitance increases due to the dielectric material and so the stored energy increases because Note that because the potential difference doesn‘t change the electric field must remain the same. Hence more charge must flow into the capacitor and thus the capacitance increases.
U0 = C0V02 / 2
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U = CV02 / 2 =κC0V0
2 / 2 >U0
CQ: Induced Charges
A point charge +Q is placed at the center of the two conducting cylinders. The induced charges on the surfaces on the cylinders are:
1. Q(I1) = Q(I2) = -Q; Q(O1) = Q(O2)= +Q 2. Q(I1) = Q(I2) = +Q; Q(O1) = Q(O2)= -Q 3. Q(I1) = -Q; Q(O1) = +Q; Q(I2) = Q(O2)= 0 4. Q(I1) = -Q; Q(O2)= +Q; Q(O1) = Q(I2)= 0 69
outer (O2)
outer (O1)
inner (I2)
inner (I1)
+Q+
CQ Ans.: Induced Charges Answer: 1. The inner faces are negative, the outer faces are positive. The surface I1 must have a charge –Q. The total charge on each cylinder is zero, therefore O1 has charge +Q. This requires I2 to have charge –Q and hence O2 has charge +Q.
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outer (O2)
outer (O1)
inner (I2)
inner (I1)
+Q+
+Q
+Q
Q
Q
CQ: Sign of Charge Suppose you insert a charge Q into the center of the Ice Pail. Connect the red lead of the voltage sensor to the inner conductor and the black lead to the outer conductor. Suppose you measure a negative voltage difference V = V1 – V2 < 0. The sign of the inserted charge is 1. negative. 2. positive. 3. zero.
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outer (O2)
outer (O1)
inner (I2)
inner (I1)
Q = ?
V
CQ Ans.: Sign of Charge Answer 1. Because the voltage difference is negative, V = V1 – V2 < 0, the sign of the charge on O1 is negative. Therefore the sign of the charge on I1 is positive, hence the sign of the inserted charge on the charge probe is negative, Q < 0. 1. negative. 2. positive. 3. zero.
72 outer (O2)
outer (O1)
inner (I2)
inner (I1)
Q < 0
V