CH 27 - Gauss's Law
If the electric flux through a closed surface is zero, the electric field at points on that surface must be zero. True or False? What equation do you use to find the electric field outside a closed surface?
False. The electric field would be at its maximum at points on the surface. You would calculate it using E = k*Q / r2
Describe what electric potential is. How does it differ from electric potential energy?
Electric potential is a scalar quantity and the voltage between two points. You might increase electric potential by moving a negative charge in the direction of a positive electric field.
Vr = kq/r
How much energy does it take to bring a charge q0 from infinity to a distance r from a point charge q?
Delta U = q0 *Vr
What do capacitors do? What is the electric potential inside a parallel plate capacitor?
Capacitors store energy. Vp = E*s where s is distance from the negative plate.
At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what distance from the line will the field strength to be 2000 N/C?
A) 2D
B) 2D
C) D/ 2
D) D/2
E) D/4
D) D/2
Using equation E = 2*k*lambda/ r we can see that in order to double electric field strength you would have to multiply (2*k*lambda/ r) by 2 (or divide by 1/2)
A negative charge, if free, will tend to move
A) from high potential to low potential.
B) from low potential to high potential.
C) toward infinity.
D) away from infinity.
E) in the direction of the electric field.
B) from low potential to high potential.
As discussed in lecture, negative charges will move from low equipotential surfaces to high equipotential surfaces.
Suppose you have two point charges of opposite sign. As you move them farther and farther apart, the potential energy of this system relative to infinity...
A) increases.
B) decreases.
C) stays the same.
How do you know this?
A) increases.
The two forces are attracted to each other and as you separate them, their potential energy increases because you are adding work to separate them from each other.
What is the definition of a battery and what is the equation to find the voltage across one? How would you calculate voltage across two batteries in series?
Batteries contain electrochemical cells and convert stored chemical energy into electrical energy.
delta V = Wchemical/q = ε
delta Vseries = delta V1 + delta V2
If a rectangular area is rotated in a uniform electric field from the position where the maximum electric flux goes through it to an orientation where only half the flux goes through it, what has been the angle of rotation?
A) 45°
B) 26.6°
C) 90°
D) 30°
E) 60°
E) 60°
Using the equation Flux = E*A*cos(theta)
Think about when cos(theta) would be equal to 1/2
Respond True or False to the following statements and explain your reasoning:
1) If the electrical potential in a region is constant, the electric field must be zero everywhere in that region.
2) If the electric field is zero everywhere inside a region of space, the potential must also be zero in that region.
3) When the electric field is zero at a point, the potential must also be zero there.
1) True. E is essentially the derivative of V taking the derivative of both sides of the equation VB - VA = -Integral (E*ds). When V is constant, slope of V is zero, and therefore E is zero.
2) False. We know V = k*q/r so inside a closed charged surface, V does not = 0 and E = 0 because the equation is E = k*q / r2 is only valid at/outside the surface of the closed surface.
3) False. Looking at answers to 1) and 2) we know this is false for similar reasons. If E is zero, V must be constant but it does NOT have to also be zero.
Which statements are true for an electron moving in the direction of an electric field? (There may be more than one correct choice.)
A) Its electric potential energy increases as it goes from high to low potential.
B) Its electric potential energy decreases as it goes from high to low potential.
C) Its potential energy increases as its kinetic energy decreases.
D) Its kinetic energy decreases as it moves in the direction of the electric field.
E) Its kinetic energy increases as it moves in the direction of the electric field.
How do you know these answers are correct?
A), C), and D)
A) Its electric potential energy increases as it goes from high to low potential.
We know this is true because electrons move in the opposite direction of an electric field, so we are applying work to push it in the same direction of the electric field.
C) Its potential energy increases as its kinetic energy decreases.
Because of the conservation of energy, we know that since the potential energy is increasing as we move the electron in the direction of the electric field, kinetic energy must decrease.
D) Its kinetic energy decreases as it moves in the direction of the electric field.
Same reason as C)
What do Van de Graaff generators do? Have you every seen one used in person?
A Van de Graaff generator separates charges and accumulates positive charge in the metal sphere at the top.
I have experimented with one of these in my high school AP Physics class. We made our hair stand up.
A charge of 1.0 × 10-6 μC is located inside a sphere, 1.25 cm from its center. What is the electric flux through the sphere due to this charge? (ε0 = 8.85 × 10-12 C2/N · m2)
0.11 N * m2/C
Flux = E*A
E = k*Q/ r2 and k = 1/(4*pi*ε0)
Flux = Q/ε0 ... because 4*pi*r2 cancel out
Flux = (1.0 × 10-12 C) / (8.85 × 10-12 C2/N · m2) = 0.11 N * m2/C
A sphere with radius 2.0 mm carries +1.0 μC of charge distributed uniformly throughout its volume. What is the potential difference, VB - VA, between point B, which is 4.0 m from the center of the sphere, and point A, which is 9.0 m from the center of the sphere? (k = 1/4πε0 =8.99 × 109 N · m2/C2)
-->1248.6 V
Potential Difference Equation:
VB-VA = k*q*(1/rA - 1/rB)
Plugging in q = 1.0μC and rA=4m and rB=9m and k = 8.99*109N*m2/C2
VB-VA = 1248.6 V
Two point charges of +1.0 μC and ‐2.0 μC are located 0.50 m apart. What is the minimum amount of work needed to move the charges apart to double the distance between them? ( k = 1/4πε0 = 8.99 × 109 N · m2/C2)
+18 mJ
We know electric potential energy is the amount work applied to an object/point charge. U = W = qV. We also know r = 0.5m and r' = 1m
V = kq / r
W = k*q1*q2/r
W' = k*q1*q2/r'
Work done = k*q1*q2*(1/r - 1/r')
Plugging in values, W = 18 * 10-3 J
What is Kirchhoff's Loop Law? What kind of problems might you use it in?
delta Vclosed loop = Sum (delta V) = 0
delta V = W/q
This Law is heavily applicable when analyzing circuits. If you know certain voltages you can use this law to help find the others.
Two concentric conducting spherical shells produce a radially outward electric field of magnitude 49,000 N/C at a point 4.10 m from the center of the shells. The outer surface of the larger shell has a radius of 3.75 m. If the inner shell contains an excess charge of ‐5.30 μC, find the amount of charge on the outer surface of the larger shell. (k = 1/4πε0 = 8.99 × 109 N ·m2/C2)
91.6 μC
Use equation E = k*q/ r2
Rearrange q = E*r2 / k
Two point charges of +2.0 μC and ‐6.0 μC are located on the x-axis at x = ‐1.0 cm and x = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero? (k = 1/4πε0 = 8.99 × 109 N · m2/C2)
r = 0.03m = 3cm
V = k*q/r
Due to 2.0 μC charge: V2 at origin = 1.8*106 V
Due to -6.0 μC charge: V-6 at origin = -2.7*106 V
V2+V-6+V3 = 0
V3 = (8.99 × 109)*(3.0*10-6)/r
(1.8*106 V)+(-2.7*106 V)+(8.99 × 109)*(3.0*10-6)/r) = 0
Solve for r.. r = 0.03m
If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed? (e = 1.60 × 10-19 C, k = 1/4πε0 = 8.99 × 109 N · m2/C2, me = 9.11 x 10-31 kg)
--> 5.9 × 107 m/s
Conservation of Energy means U = KE
q*Vr= 1/2 *m*v2
v = sqrt(2*q*Vr/m)
Plugging in Values,
v = 5.9 × 107 m/s
What is the potential difference between xi=20cm and xf=40cm in the uniform electric field Ex=1000V/m?
Vf - Vi = -200V
E = -delta V/ delta x
Vf - Vi = -E (xf - xi)
=- 1000 V/m * (0.4-0.2)m
= -200 V