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
zaharov [31]
3 years ago
5

Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q

2 = +1.00 µC, and negligible thicknesses. What is the magnitude of the electric field E at radial distance (a) r = 4.00 m, (b) r =0.700 m, and (c) r = 0.200 m? With V = 0 at infinity, what is V at (d) r = 4.00 m, (e) r = 1.00 m, (f) r = 0.700 m, (g) r = 0.500 m, (h) r = 0.200 m, and (i) r = 0? (j) Plot the E(r) and V(r) dependencies.

Physics
1 answer:
scZoUnD [109]3 years ago
8 0

Complete Question

The diagram for this question is shown on the first uploaded image  

Answer:

a E =1.685*10^3 N/C

b E =36.69*10^3 N/C

c E = 0 N/C

d V = 6.7*10^3 V

e   V = 26.79*10^3V

f   V = 34.67 *10^3 V

g   V= 44.95*10^3 V

h    V= 44.95*10^3 V

i    V= 44.95*10^3 V

Explanation:

From the question we are given that

       The first charge q_1 = 2.00 \mu C = 2.00*10^{-6} C

       The second charge q_2 =1.00 \muC = 1.00*10^{-6}

      The first radius R_1 = 0.500m

      The second radius R_2 = 1.00m

 Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}

And Potential \ Difference = \frac{1}{4\pi \epsilon_0}   [\frac{q_1 }{r}+\frac{q_2}{R_2} ]

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

                      r  = 4.00 m

           E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}

                = 1.685*10^3 N/C

Considering b

           r = 0.700 m \ , R_2 > r > R_1

This implies that the electric field would be

            E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}

             This because it the electric filed of the charge which is below it in distance that it would feel

            E = 8*99*10^9  \frac{2*10^{-6}}{0.4900}

               = 36.69*10^3 N/C

   Considering c

                      r  = 0.200 m

=>   r

 The electric field = 0

     This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

       Considering d

                  r  = 4.00 m

=> r > R_1 >r>R_2

Now the potential difference is

                  V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V

This so because the distance between the charge we are considering is further than the two charges given  

          Considering e

                       r = 1.00 m R_2 = r > R_1

                V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V

          Considering f

              r = 0.700 m \ , R_2 > r > R_1

                      V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V

          Considering g

             r =0.500\m , R_1 >r =R_1

   V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

          Considering h

                r =0.200\m , R_1 >R_1>r

  V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

           Considering i    

   r =0\ m \ , R_1 >R_1>r

  V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

You might be interested in
A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el
Irina-Kira [14]

Answer:

71.19 C

Explanation:

25C = 25 + 273 = 298 K

Applying the ideal gas equation we have

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:

T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C

4 0
3 years ago
A sailboat moves north for a distance of 10.00 km when blown by a wind from the exact south with a force of 5.00 x 10^4 n. how m
Nataliya [291]
Work is defined as the force times the distance which is mathematically expressed W = Fxd. The given force is 5x10^4 and the distance is 10000 m (the distance is converted as meter because Nm = J) the work done by the wind is W = 5x10^4 N (10000) = 500 x 10^6 Joules. I hope it answered your question
8 0
3 years ago
If the same types of fossils are found in two separate rock layers, it's likely that the two rock layers ___
valentina_108 [34]
Are part of one continuous deposit
6 0
3 years ago
What specific structures are in electroreceptors?
Slav-nsk [51]
Electroreception is limited to aquatic environments because on here is the resistivity of the medium is low enough for electric currents to be generated as the result of electric fields of biological origin.  In air, the resistivity of the environment is so high that electric fields from biological sources do not generate a significant electric current.  Electroreceptor are found in a number of species of fish,  and in at least one species of mammal,  the Duck-Billed platypus.
7 0
3 years ago
in a human cannonball a person is shot from a cannon with a barrel that is 3.05 m long. 16700 J of work are done to accelerate t
ankoles [38]
The work done to accelerate the acrobat is given by
W=Fd
where F is the force applied and d the distance of application of the force.
If the barrel is 3.05 m long, then d=3.05 m. Therefore we can find the force:
F= \frac{W}{d} = \frac{17600 J}{3.05 m} =5475 N
6 0
3 years ago
Other questions:
  • Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
    14·1 answer
  • Dolphin echolocation is similar to ultrasound. Reflected sound waves
    9·1 answer
  • A ball rolls off the edge of a table with a fairly large horizontal velocity. Which of the following statements are true? (Selec
    5·1 answer
  • Can anyone solve this please.........
    14·1 answer
  • Which is a correct step in a scientific experiment involving acids and bases?
    6·1 answer
  • A cyclist rode at an average speed of 15mph for 30 miles how long was the ride
    15·1 answer
  • Current that moves in one direction from negative to positive. May be created by a battery. Is generally NOT found in U.S. Elect
    12·2 answers
  • 19 point please please answer right need help
    6·1 answer
  • What determines the state of matter for any
    12·2 answers
  • Q7, how the ans is 4.9 not 19.6?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!