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Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q2 = +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.

1 Answer

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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

Step-by-step 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 = (1)/(4\pi\epsilon_0)(q_1+\ q_2)/(r^2)

And
Potential \ Difference = (1)/(4\pi \epsilon_0) [(q_1 )/(r)+(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 = (((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 = (1)/(4\pi \epsilon_0)(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 (2*10^(-6))/(0.4900)

=
36.69*10^3 N/C

Considering c

r = 0.200 m

=>
r<R_1<R_2

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

Considering f


r = 0.700 m \ , R_2 > r > R_1


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

Considering g


r =0.500\m , R_1 >r =R_1


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

Considering h


r =0.200\m , R_1 >R_1>r


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

Considering i


r =0\ m \ , R_1 >R_1>r


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

Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and-example-1
User Wombatp
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