A spherical metal ball is charged up with a total charge Q. The electric potential inside
the shell is:
A)
B) V = E
V =0
0
Q
C) V = k Q
R2
D) V = k Q
R
E) Don’t Know.
See explanation with Lecture 24.
Two large metal plates are separated by a distance d along the x-axis. The right hand
plate is located at d and charged up to a potential of V(d)=V0, the left hand is located at
the origin, and the plate is held at V=0.
What is the correct expression for the potential between the plates?
A) V=0 B) V=V0 C) V=V0x/d D) V=V0x2/d2
E) V=V0d/x
At x=0, V=0, at X=d, V=V0, in between the potential varies linearly.
What is the direction of the Electric field between the plates?
A) –x B) +x C) –y D) +y E) Don’t know.
A positive voltage means a positive charge, the E field goes from the positive
charge to the negative charge, so –x.
What is the correct expression for the magnitude of electric field between the plates?
A) E=0 B) E=V0 C) E=V0/d D) E=V0x/d2
E) E=V0/x
The E=dV/dx, so differentiate the previous result V=V0x/d wrt. x.
If d=3 m and V0=1 eV, what is the velocity expressed as a fraction of c, of an electron
(mass 0.5MeV/c2) that is released at the left plate?
A) 0.2 c B) 0.01c C) 0.02 c D) 0.001 c E) 0.002 c
Use: K = 12 mv 2 = U = e∆V , using units of eV, so:
1
2
( 0.5 ×10
v2
6
eV/c2 ) v 2 = 1 eV ⇒
= 4 ×10−6 ⇒ v = 2 × 10−3 c
c2
Two metal spheres, S1 whith a radius of 3 cm and S2 with a radius of 9 cm, are separated
by a large distance. They are connected by a thin conducting wire.
If a charge Q is placed on S1, what is the final (equilibrium) charge on each of the
spheres?
A) Q1=Q/2, Q2=Q/2
B) Q1=Q/4, Q2=3Q/4
C) Q1=3Q/4, Q2=Q/4
D) Q1=Q, Q2=0
E) Don’t know.
The electric potential on both spheres must be the same (they are connected
by a conducting wire!). Potential of S1 is V1=k Q1/R1, S2: V2=k Q2/R2, so,
since we have V1=V2, Q1/R1=Q2/R2. We also have conservation of charge, so
Q1+Q2=Q. Two equations, two unknowns, solve for Q1 and Q2.