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Please Take a look at the attached screenshots to see the questions

please make use of the solutions for clarification!!

1.

E

E

E

E

E

E

E

E

A hypothetical closed surface is

made by attaching a smoothly dis-

torted hemispherical surface (2) of

radius “R” to a flat circular base

plate (1) also of radius “R” as shown

in cross section at left. The closed

surface is immersed in a constant

horizontal electric field (E) and it

Ã¢â‚¬â€ encloses no charge.

E

E

E

E

0

2

E

E

E

a. What is the flux of the (E) through the closed surface?

b. What is the flux of the (E) through the back surface (1)?

c. What is the flux of the (E) through the front surface (2)?

(Hint: make use of 0.

closed surface (0)zuface panels)

2.

ty 4

3

R

2

A hypothetical infinitely long (i.e.

L 00) closed semicircular cylindri-

cal shell surface has its axis aligned

on the (z) axis ; the shell has an in-

ner radius of “r” and an outer ra-

dius of “R”. Note that we can as-

sume the shell to be enclosed by

+X four surfaces (1,2,3 and 4) as indica-

ted in the figure. An infinite line of

charge (+ ) runs along the (z) axis

thus coinciding with shell’s axis.

)

4

Note that in this problem you will have to use the inte-

gral definition of flux for each surface:

Q = SE-nda

where “M” the outward unit normal vector.

a. What is the flux of the electric field through each of the four

enclosing “surface panels” (i.e. the surfaces 1, 2, 3 and 4)?

b. Using the fact that closed surface

3(0)surface panels

find

the flux of the electric field through the shell.

Use Gauss’s Law to redo part b.

C.

3.

R

The figure at left shows a cross sec-

tion of a solid nonconducting sphere

(radius = R = 0.80 m) which has a

a total positive charge (Q) of uniform-

ly distributed throughout its volume

such that it has a constant charge

per unit volume of p = + 64 ucoul/m3.

ÃÂ

a. What is the value of the charge (Q) in the sphere?

b. Derive an expression (in terms of p) for the magnitude of the

electric field in the region Osr R and then calculate its value

at the point r = 5.0 m.

e. Referring to part d, convert your expression for (E) from one

in terms of (o) to one in terms of (Q).

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