1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
solniwko [45]
3 years ago
12

A sphere of radius R = 0.295 m and uniform charge density -151 nC/m^3 lies at the center of a spherical, conducting shell of inn

er and outer radii 3.50 R and 4.00 R , respectively. If the conducting shell carries a total charge of Q = 66.7nC , find the magnitude of the electric field at the at the following radial distances from the center of the charge distribution.
a. 0.760R
b. 3.90R
c. 2.80R
d. 7.30R
Physics
1 answer:
cupoosta [38]3 years ago
6 0

Answer:

a) -1.27*10³ N/C b) 0 c) -0.21*10³ N/C d) 0.1*10³ N/C

Explanation:

a) r = 0.76R

As this distance is inside the sphere, we need to know how much charge is enclosed within this distance for the center, as follows:

Q = ρ*V(r) = ρ*\frac{4}{3} *\pi *r^{3}

where r = 0.760* R = 0.760* 0.295 m = 0.224 m, and ρ = -151 nC/m³

Q = -151e-9 *\frac{4}{3} *\pi *0.224m^{3} = -7.11e-9 C

Applying Gauss' Law to a spherical gaussian surface of r= 0.76R, as the electric field is radial, and directed inward, we can write the following equation:

E*A = Q/ε₀, where Q= -7.11 nC, A= 4*π*(0.76R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{-7.11e-9C}{(0.76*0.295m)^{2}} =-1.27e3 N/C

⇒ E = -1.27*10³ N/C

b) r= 3.90 R

As this distance falls inside the conducting shell, and no electric field can exist within a conductor in electrostatic condition, E=0

c) r = 2.8 R

As this distance falls between the sphere and the inner radius of the shell, we can calculate the electric field, applying Gauss' law to a gaussian surface of radius equal to r= 2.80 R.

First we need to find the total charge of the sphere, as follows:

Q = ρ*V =

Q = -151e-9 *\frac{4}{3} *\pi *0.295m^{3} = -16.2e-9 C

In the same way that for a) we can write the following expression:

E*A = Q/ε₀, where Q= -16.2 nC, A= 4*π*(2.8R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{-16.2e-9C}{(2.8*0.295m)^{2}} =-0.21e3 N/C

⇒ E = -0.21*10³ N/C

d) r= 7.30 R

In order to find the electric field at this distance, which falls beyond the outer radius of the shell, we need to find the total charge on the outer surface.

As the sphere has a charge of -16.2 nC, and the total charge of the conducting shell is 66.7nC, in order to make E=0 inside the shell, the total charge enclosed by a gaussian surface with a radius larger than the inner radius of the shell and shorter than the outer one, must be zero, which means that a charge of +16.2 nC must be distributed on the inner surface of the shell.

This leaves an excess charge on the outer surface of the shell as follows:

Qsh = 66.7 nC - 16.2 nC = 50.5 nC

Now, we can repeat the same process than for a) and c) as follows:

E*A = Q/ε₀, where Q= 50.5 nC, A= 4*π*(7.3R)² and ε₀ =8.85*10⁻¹² C²/N*m²

We can solve for E, as follows:

E = \frac{1}{4*\pi*8.85e-12C2/N*m2 } *\frac{50.5e-9C}{(7.3*0.295m)^{2}} =0.1e3 N/C

⇒ E = 0.1*10⁻³ N/C

You might be interested in
Where does denitrification happen?
Alchen [17]
The correct anwser is in deep soil or anoxic aquatic sediment.
6 0
3 years ago
Which of the following types of radiation can penetrate through paper but not through wood?
andrew-mc [135]
The type of radiation that can penetrate through paper, but not through wood is called beta rays. Beta rays can penetrate paper and air, but a thin piece of alimony can stop it. Gamma can cut through anything except lead and many inches of concrete. Alpha can be stopped by paper and not penetrated. The correct answer is B.
4 0
3 years ago
Read 2 more answers
The mole is 6.02 x 10 23 particles. If a person masses out the correct molar mass in grams for a substance then she would have a
Leto [7]

Answer:

1 mole of H2O is 18 grams (2 g H + 16 g Oxygen)

36 / 18 = 2

So 2 moles = 2 * 6.02E23 = 12.04E23 = 1.204E24

7 0
3 years ago
MRU(movimiento rectilineo uniforme) un movil viaja en linea recta con una velocidad media de 12metros /segundos durante 9segundo
liubo4ka [24]

Answer:

a) 141.6m

b) 8.4m/s

Explanation:

a) to find the total displacement you use the following formula for each trajectory. Next you sum the results:

x_1=v_1t_1=(12m/s)(9s)=108m\\\\x_2=v_2t_2=(4.8m/s)(7s)=33.6m\\\\x_T=x_1+x_2=108m+33.6m=141.6m

hence, the total distance is 141.6m

b) the mean velocity of the total trajectory is given by:

v_m=\frac{v_1+v_2}{2}=\frac{12m/s+4.8m/s}{2}=8.4\frac{m}{s}

hence, the mean velocity is 8.4 m/s

8 0
3 years ago
Help pleasee
agasfer [191]

Answer:

i/f = i/o + i/i       f = focal, o = object, i = image

1 / i = 1 / f - 1 / o  =    (o - f) / o f

i = o * f / ( o - f)      image distance

i = 12.5 * 22 / (12.5 - 22) = -28.9 cm

Image is real

Image is 28.9 cm to left of lens

M = - i / o = = 28.9 / 12.5 = 2.3     magnification (convex lens)

8 0
2 years ago
Other questions:
  • What’s the difference between weight and support force? *oof*
    11·1 answer
  • Two or more atoms can be held together through shared
    15·1 answer
  • What season is it when the earth axis is tilted so that it is facing toward the sun?
    6·2 answers
  • The work done when a force moves a body through a distance of 15m is 1800j. What is the value of the force applied
    7·1 answer
  • In much the same way, the sun warms you by radiation on a clear day. The space between the sun and Earth has no air to transfer
    12·1 answer
  • Life cycle of a medium mass star
    7·1 answer
  • Which graph accurately shows the relationship between kinetic energy and mass as mass increases
    6·1 answer
  • What is the mass of a vehicle that has 50,000 N and 25M/S/S
    9·1 answer
  • Bob Beamon's 1968 Olympic long jump set a world record which remains unbroken to this day. This amazing jump resulted from an in
    14·2 answers
  • A body is projected vertically upwards with a speed 100 m/s from the ground then distance it covers in its last second of its up
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!