week 2

In a uniform electric field in empty space, a 4 C charge is placed and it feels an electrical force of 12 N. If this charge is removed and a 6 C charge is placed at that point instead, what force will it feel? 1) 12 N 2) 8 N 3) 24 N 4) no force 5) 18 N Q

In a uniform electric field in empty space, a 4 C charge is placed and it feels an electrical force of 12 N. If this charge is removed and a 6 C charge is placed at that point instead, what force will it feel? 1) 12 N 2) 8 N 3) 24 N 4) no force 5) 18 N Since the 4 C charge feels a force, there must be an electric field present, with magnitude: E = F / q = 12 N / 4 C = 3 N/C Once the 4 C charge is replaced with a 6 C Q charge, this new charge will feel a force of: F = q E = (6 C)(3 N/C) = 18 N Follow-up: What if the charge is placed at a different position in the field?

Which electric field is responsible for the trajectory of the proton? 1. 2. 3. 4. 5.

Which electric field is responsible for the trajectory of the proton? Parabolic trajectories occur when there is constant acceleration. This requires a uniform electric force (compare to projectile motion, but here the force of gravity points to the right) 1. 2. 3. 4. 5.

An electron is launched at a 45 degree angle and a speed of 5.0 x 10 6 m/s from the positive plate of a parallel-plate capacitor as shown in the figure. The electron lands 4.0 cm away. (a) What is the electric field strength inside the capacitor? (b) What is the smallest possible spacing between the plates? (me=9.1 x 10-31 kg, e=1.6 x 10-19 C) Ans: (a) 3.55 kn/c (b) 1.0 cm

A proton orbits a 1.0 cm-diameter metal ball 1.0 mm above the surface. The orbital period is 1.0 μs. What is the charge on the ball. (mp = 1.67 x 10-27 kg, e=1.6 x 10-19 C) Ans: 9.9 nc Q = 1 K m e 4π 2 R 3 T 2

1. Left. At the position of the dot, the electric field points 2. Down. 3. Right. 4. Up. 5. The electric field is zero.

1. Left. At the position of the dot, the electric field points 2. Down. 3. Right. 4. Up. 5. The electric field is zero.

What is the direction of - Q + Q the electric field at the position of the X? 2 3 + Q 1 5 4

What is the direction of - Q + Q the electric field at the position of the X? 2 3 + Q 1 5 4 The two +Q charges give a resultant E field that is down and to the right. The Q charge has an E field up and to the left, but smaller in magnitude. Therefore, the total electric field is down and to the right. Follow-up: What if all three charges reversed their signs?

Field of Electric Dipole: 1. along axis O H + H + + +q + E = E + + E d p q dipole moment: p dq dipole aligned with x-axis, compute field on points on axis (COB) E(x, 0, 0) = 1 2πɛ 0 p x 3 (x d)

Two charges are fixed along the x-axis. They produce an electric field E directed along the negative y-axis at the indicated point. Which of the following is true? 1) charges are equal and positive 2) charges are equal and negative 3) charges are equal and opposite 4) charges are equal, but sign is undetermined 5) charges cannot be equal y E Q 1 Q 2 x

Two charges are fixed along the x-axis. They produce an electric field E directed along the negative y-axis at the indicated point. Which of the following is true? 1) charges are equal and positive 2) charges are equal and negative 3) charges are equal and opposite 4) charges are equal, but sign is undetermined 5) charges cannot be equal The way to get the resultant PINK vector y is to use the GREEN and BLUE vectors. These E vectors correspond to equal charges (because the lengths are equal) E that are both negative (because their directions are toward the charges). Q Q 1 2 Follow-up: How would you get the E field to point toward the right? x

Field of Electric Dipole: 2. bisecting plane O H + H + + y E = E + + E y q x = a E r θ E + E cos θ = a r r + q x = a x or (COB) E = 1 4πɛ 0 p E(0, y, z) = 1 4πɛ 0 r 3 p ( y 2 + z 2 ) 3 ( y 2 + z 2 a)

Electric Field Lines / Equipotential Surfaces (applet)

+ + q 2q q (a) (b) (c) q + + q q + +q q/2 + q (d) (e) (f)

The number of electric field lines emerging from any closed surface is proportional to the charge enclosed (this will be the central notion next week: Gauss s Law) 4 1 1 4 1 4 + + q 2 2q q 2 2 3 3 3 (a) (b) (c) 4 2 1 2 1 q + + q q 2 + +q q/2 + 1 4 3 3 q 3 (d) (e) (f)

What are the signs of the charges whose electric fields are shown at right? 1) 2) 3) 4) 5) no way to tell

What are the signs of the charges whose electric fields are shown at right? 1) 2) 3) 4) 5) no way to tell Electric field lines originate on positive charges and terminate on negative charges.

Which of the charges has the greater magnitude? 1) 2) 3) Both the same

Which of the charges has the greater magnitude? 1) 2) 3) Both the same The field lines are denser around the red charge, so the red one has the greater magnitude. Follow-up: What is the red/green ratio of magnitudes for the two charges?

Uniformly charged ring, along axis dq a O x x 2 + a 2 θ P de θ de de x dq (COB) E(x, 0, 0) = 1 4πɛ 0 xq (x 2 + a 2 ) 3/2 î E(x, 0, 0) 1 4πɛ 0 x x 3 î E(x, 0, 0) 1 4πɛ 0 x a 3 î x a x a

A piece of plastic is uniformly charged with surface charge density η1. The plastic is then broken into a large piece with surface charge density η2 and a small piece with surface charge density η3. Rank in order, from largest to smallest, the surface charge densities η1 to η3. 1. η 2 = η 3 > η 1 2. η1 > η2 > η3 3. η 1 > η 2 = η 3 4. η 3 > η 2 > η 1 5. η 1 = η 2 = η 3

A piece of plastic is uniformly charged with surface charge density η1. The plastic is then broken into a large piece with surface charge density η2 and a small piece with surface charge density η3. Rank in order, from largest to smallest, the surface charge densities η1 to η3. 1. η 2 = η 3 > η 1 2. η1 > η2 > η3 3. η 1 > η 2 = η 3 4. η 3 > η 2 > η 1 5. η 1 = η 2 = η 3

Which of the following actions will increase the electric field strength at the position of the dot? 1. Make the rod longer without changing the charge. 2. Make the rod shorter without changing the charge. 3. Make the rod fatter without changing the charge. 4. Make the rod narrower without changing the charge. 5. Remove charge from the rod.

Which of the following actions will increase the electric field strength at the position of the dot? 1. Make the rod longer without changing the charge. 2. Make the rod shorter without changing the charge. 3. Make the rod fatter without changing the charge. 4. Make the rod narrower without changing the charge. 5. Remove charge from the rod.

uniformly charged disk, field on axis dr r R x σ = Q πr 2 (COB) E x = 2kQ R 2 [ 1 x x2 + R 2 ] = 2πkσ [ 1 ] x x2 + R 2 limits x R E x = kq x 2 x R (R ) (infinite plane) E x = 2πkσ

Rank in order, from largest to smallest, the electric field strengths E a to E e at these five points near a plane of charge. 1. E a = E b = E c = E d = E e 2. E a > E c > E b > E e > E d 3. E b = E c = E d = E e > E a 4. E a > E b = E c > E d = E e 5. E e > E d > E c > E b > E a

Rank in order, from largest to smallest, the electric field strengths E a to E e at these five points near a plane of charge. 1. E a = E b = E c = E d = E e 2. E a > E c > E b > E e > E d 3. E b = E c = E d = E e > E a 4. E a > E b = E c > E d = E e 5. E e > E d > E c > E b > E a

Rank in order, from largest to smallest, the forces F a to F e a proton would experience if placed at points a e in this parallel-plate capacitor. 1. F a = F b = F c = F d = F e 2. F a = F b > F c > F d = F e 3. F a = F b = F d = F e > F c 4. F e > F d > F c > F b > F a 5. F e = F d > F c > F a = F b

Rank in order, from largest to smallest, the forces F a to F e a proton would experience if placed at points a e in this parallel-plate capacitor. 1. F a = F b = F c = F d = F e 2. F a = F b > F c > F d = F e 3. F a = F b = F d = F e > F c 4. F e > F d > F c > F b > F a 5. F e = F d > F c > F a = F b

challenge: uniformly charged cylinder, field on axis recall dr R E x = 2kQ R 2 [ 1 ] [ x = 2πkσ 1 x2 + R 2 ] x x2 + R 2 r x use ρ = Q πr 2 L de x = 2πkρdx [ 1 ] d x (d x)2 + R 2 E x = 2πkρ [L + (L/2 x) 2 + R 2 (L/2 + x) 2 + R 2 ] E x kρπr2 L x 2 x R, L