Question

We want to find the electric field near an infinite charged conducting plane. Gauss' Law works for any closed surface whatsoever. The art of using it is to match the shape of the Gaussian surface to the symmetry of the problem. The figure shows a Gaussian surface of the variety called a "pillbox," a cylinder with an arbitrary area A on each end and extending for a height h from the surface. The conductor bears a surface charge density of ? in [C/m²]. The left end of the pillbox is buried inside the conductor. By symmetry, the electric field E must be perpendicular to the surface (read that again, and look at the diagram). Always start by writing down Gauss' Law: ? E ? dA = q_enclosed / ?_0 The left hand side (lhs) is an integral over the closed surface of the pillbox. Break this into the sum of three integrals: one each for the left end, the right end, and the sides of the cylinder: ? total = ? left end + ? right end + ? sides where only the first is a closed integral. These three pieces make up the closed surface. Now evaluate each of these integrals separately: Left end: ? E ? dA = 0 (easy because E = 0 inside a conductor). Right end: ? E ? dA = (what is the angle between E and A on the right end cap?). Sides: ? E ? dA = (what is the angle between E and A on the right end cap? 90°, right?). Sum these to get: ? E ? dA = Now for the right hand side (rhs). You need the charge enclosed. But all of the charge resides on the surface of the conductor. What is the area of the surface inside the pillbox? How do you get from area A in [m²] to charge Q in [C] using the information you have? (Hint: it's in the first paragraph.) Q_enclosed = ________. Now set the lhs equal to the rhs (divided by ?_0), simplify, and you are done. Put the answer here: EA = therefore E = Notice a funny thing about this formula. It does not involve the distance from the surface. Why do you suppose this is? What assumption was made that leads to this, and when will it break down?

          We want to find the electric field near an infinite charged conducting
plane. Gauss' Law works for any closed surface whatsoever. The art of
using it is to match the shape of the Gaussian surface to the symmetry of
the problem. The figure shows a Gaussian surface of the variety called a
"pillbox," a cylinder with an arbitrary area A on each end and extending
for a height h from the surface. The conductor bears a surface charge
density of ? in [C/m²]. The left end of the pillbox is buried inside the
conductor. By symmetry, the electric field E must be perpendicular to the
surface (read that again, and look at the diagram).

Always start by writing down Gauss' Law:
? E ? dA = q_enclosed / ?_0

The left hand side (lhs) is an integral over the closed surface of the
pillbox. Break this into the sum of three integrals: one each for the left
end, the right end, and the sides of the cylinder:
? total = ? left end + ? right end + ? sides

where only the first is a closed integral. These three pieces make up the
closed surface. Now evaluate each of these integrals separately:

Left end: ? E ? dA = 0 (easy because E = 0 inside a conductor).

Right end: ? E ? dA = ________ (what is the angle between E and A on the right end cap?).

Sides: ? E ? dA = ________ (what is the angle between E and A on the right end cap? 90°, right?).

Sum these to get: ? E ? dA = ________

Now for the right hand side (rhs). You need the charge enclosed. But all of the charge resides on the
surface of the conductor. What is the area of the surface inside the pillbox? How do you get from area A
in [m²] to charge Q in [C] using the information you have? (Hint: it's in the first paragraph.)

Q_enclosed = ________.

Now set the lhs equal to the rhs (divided by ?_0), simplify, and you are done. Put the answer here:

EA = ________ therefore E = ________

Notice a funny thing about this formula. It does not involve the distance from the surface. Why do you
suppose this is? What assumption was made that leads to this, and when will it break down?

We want to find the electric field near an infinite charged conducting
plane. Gauss' Law works for any closed surface whatsoever. The art of
using it is to match the shape of the Gaussian surface to the symmetry of
the problem. The figure shows a Gaussian surface of the variety called a
"pillbox," a cylinder with an arbitrary area A on each end and extending
for a height h from the surface. The conductor bears a surface charge
density of ? in [C/m²]. The left end of the pillbox is buried inside the
conductor. By symmetry, the electric field E must be perpendicular to the
surface (read that again, and look at the diagram).

Always start by writing down Gauss' Law:
? E ? dA = qenclosed / ?0

The left hand side (lhs) is an integral over the closed surface of the
pillbox. Break this into the sum of three integrals: one each for the left
end, the right end, and the sides of the cylinder:
? total = ? left end + ? right end + ? sides

where only the first is a closed integral. These three pieces make up the
closed surface. Now evaluate each of these integrals separately:

Left end: ? E ? dA = 0 (easy because E = 0 inside a conductor).

Right end: ? E ? dA = (what is the angle between E and A on the right end cap?).

Sides: ? E ? dA = (what is the angle between E and A on the right end cap? 90°, right?).

Sum these to get: ? E ? dA =

Now for the right hand side (rhs). You need the charge enclosed. But all of the charge resides on the
surface of the conductor. What is the area of the surface inside the pillbox? How do you get from area A
in [m²] to charge Q in [C] using the information you have? (Hint: it's in the first paragraph.)

Qenclosed = .

Now set the lhs equal to the rhs (divided by ?0), simplify, and you are done. Put the answer here:

EA = therefore E =

Notice a funny thing about this formula. It does not involve the distance from the surface. Why do you
suppose this is? What assumption was made that leads to this, and when will it break down?

Added by Patricia D.

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