Compute the potential difference AV between: a) Two conducting spherical shells of radius a and b with a charge +Q on the inner one and charge -Q on the outer one. b) Two (infinitely long) conducting cylindrical shells of radius a and b with a charge per unit length +A on the inner one and charge per unit length -1 on the outer one. c) Two (infinite) conducting sheets of charge, one with charge +o on the ry plane and with with charge parallel to the first one but at z = d. Great! Now you've done almost all the work required to understand Capacitance!

Classical Dynamics of Particles and Systems
5th Edition
ISBN:9780534408961
Author:Stephen T. Thornton, Jerry B. Marion
Publisher:Stephen T. Thornton, Jerry B. Marion
Chapter5: Gravitation
Section: Chapter Questions
Problem 5.12P
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Solve problem 6
How much work is required to assemble a uniform ball of charge with total (final) charge
Q and radius R? Hint: This is the same as the potential energy of the sphere, so use
đU = V dq and imagine "building" the sphere a layer of thickness dr at a time. Alternatively,
compute the work directly by bringing a charge dq in from infinity against the electric field
of the charge already there (and distributed as a sphere of radius r).
Problem 6.
Compute the potential difference AV between: a) Two conducting spherical shells of radius
a and b with a charge +Q on the inner one and charge -Q on the outer one. b) Two
(infinitely long) conducting cylindrical shells of radius a and b with a charge per unit length
+A on the inner one and charge per unit length -A on the outer one. c) Two (infinite)
conducting sheets of charge, one with charge +o on the ry plane and with with charge
parallel to the first one but at z =
to understand Capacitance!
d. Great! Now you've done almost all the work required
Problem 7.
Three thin conducting spherical shells have radii a < 6 < e respectively. Initially the shell
with radius a has a charge +Q and the shell with radius 6 has a charge -Q. You connect
the shells with radii a and e using a thin wire that passes through a tiny (insulated!) hole
through the middle shell and wnit for the chargo to como to
Thr m What in a)
Transcribed Image Text:How much work is required to assemble a uniform ball of charge with total (final) charge Q and radius R? Hint: This is the same as the potential energy of the sphere, so use đU = V dq and imagine "building" the sphere a layer of thickness dr at a time. Alternatively, compute the work directly by bringing a charge dq in from infinity against the electric field of the charge already there (and distributed as a sphere of radius r). Problem 6. Compute the potential difference AV between: a) Two conducting spherical shells of radius a and b with a charge +Q on the inner one and charge -Q on the outer one. b) Two (infinitely long) conducting cylindrical shells of radius a and b with a charge per unit length +A on the inner one and charge per unit length -A on the outer one. c) Two (infinite) conducting sheets of charge, one with charge +o on the ry plane and with with charge parallel to the first one but at z = to understand Capacitance! d. Great! Now you've done almost all the work required Problem 7. Three thin conducting spherical shells have radii a < 6 < e respectively. Initially the shell with radius a has a charge +Q and the shell with radius 6 has a charge -Q. You connect the shells with radii a and e using a thin wire that passes through a tiny (insulated!) hole through the middle shell and wnit for the chargo to como to Thr m What in a)
Expert Solution
Step 1

Compute the potential difference between :

  1.  Two conducting spherical shells of radius a and b with charge +Q (inner) and -Q (outer).
  2.  Two conducting cylindrical shells of radius a and b with a charge per unit length +λ (inner) and -λ (outer). 
  3.  Two conducting sheets of charge, one with charge +σ on the x-y plane and with charge -σ parallel to the first one, at z = d. 

Concepts & Formulas : 

Gauss's theorem states that "the total electric flux out of a closed surface is proportional to the electric charge enclosed by the surface".

             φ = E. ds = Qenclosedεo

Potential difference, V =-E.dl

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