Question

Two isolated, concentric, conducting spherical shells have radii R1 = 0.520 m and R2 = 1.40 m, uniform charges q1 = +2.50 μC and q2 = +2.70 μC, and negligible thicknesses. What is the magnitude of the electric field E at radial distance (a) r = 4.80 m, (b) r = 0.710 m, and (c) r = 0.210 m? With V = 0 at infinity, what is V at (d) r = 4.80 m, (e) r = 1.40 m, (f) r = 0.710 m, (g) r = 0.520 m, (h) r = 0.210 m, and (i) r = 0?

Answer 1

Answer:

Explanation:

To solve this problem we can use the Gauss' Theorem

Hence, we have:

$\int Eds=\frac{Q_N}{\epsilon_0}\\E\int ds=\frac{Q_N}{\epsilon_0}\\E=\frac{Q_N}{4\pi r^2\epsilon_0}$

where QN is the total net charge inside the Gaussian surface, r is the point where we are going to compute E and ε0 is the dielectric permitivity. For each value of r we have to take into account what is the net charge inside the Gaussian surface.

a) r=4.80m (r>R2)

QN=+2.50 μC+2.70 μC = 5.2 μC

$E=\frac{5.2*10^{-6}C}{4\pi (4.80)^2(8.854*10^{-12} \frac{C}{V m})}=2.02*10^3\frac{N}{C}$

b) r=0.70m (R1<r<R2)

QN=+2.50 μC

$E=\frac{5.2*10^{-6}C}{4\pi (0.710m)^2(8.854*10^{-12} \frac{C}{V m})}=9.53*10^4\frac{N}{C}$

c) r=0.210 (r<R1)

Inside the spherical shell of radius R1 the net charge is zero. Hence

E=0N/C

- For the calculation of the potential we have

$V=-\int \frac{Q_N}{4\pi r^2\epsilon_0}dr=\frac{Q_N}{4\pi r\epsilon_0}$

Thus, we compute the potential by using the net charge of the Gaussian surface

d) r=0.210 (r<R1)

Inside the spherical shell the net charge is zero, thus

E=0N/C

e) r=1.40m (R1<r<=R2)

In this case we take the net charge from the first spherical shell

QN=+2.50 μC

$V=\frac{2.50*10^{-6}C}{4\pi (1.4m)(8.854*10^{-12}\frac{C}{Vm})}=1.604*10^{4}\frac{Nm}{C}$

f) r=0.70m

QN=+2.50 μC

V=3.164*10^{4}Nm/C

g) r=0.52

QN=0

V=0

h) r=0.2

QN=0

V=0

HOPE THIS HELPS!!

Answer 2

Answer:

A) E = 2031.7 N/C

B) E = 44,643.8 N/C

C) E = 0 N/C

D) V = 9752.14 N/C

E) V = 33,435.91 N/C

F) V = 49,058 N/C

G) V = 60,639.65 N/C

H) V = 60,639.65 N/C

I) V = 60,639.65 N/C

Explanation:

We shall make use of the following;

At r < R1; E_in = 0 - - - - - (eq1)

At R1 < r < R2 ;

E_bet = q1/(4πε_o(r²)) -- - - - - (eq2)

At R1 < r < R2 ;

E_out = (q1 + q2)/(4πε_o(r²)) - (eq3)

At R1 < r < R2 ;

V_in = [1/(4πε_o)]•[(q1/R1) + (q2/R2)] - - - - - (eq4)

At R1 < r < R2 ;

V_bet = [1/(4πε_o]•[(q1/r) + (q2/R2)] - - - - - (eq5)

At r > R2 ;

V_out= (q1 + q2)/(4πε_o(r)) - - - - - (eq6)

A) r = 4.80 m. This falls into r > R2. But since we are looking for electric field, E, we will use eq (3)

Thus, E = (q1 + q2)/(4πε_o(r)²)

q1 = +2.50 μC = 2.5 x 10^(-6) C

q2 = 2.7 μC = 2.7 x 10^(-6) C

ε_o = 8.84 x 10^(-12) F/m

E = [(2.5 x 10^(-6))] + (2.7 x 10^(-6))] /(4πx8.84 x 10^(-12)x(4.8)²)

= 2031.7 N/C

B) r = 0.710 m, we'll use eq (2)

E_bet = q1/(4πε_o(r²))

E_bet = [2.5 x 10^(-6)]/(4πx8.84 x 10^(-12)x(0.71²)) = 44,643.8 N/C

C) r = 0.210 m

This corresponds with condition for eq 1 where enclosed charge is zero.

Thus, E_in = 0 N/C

D) V at r = 4.80 m

Corresponds to eq(6) where;

V_out= (q1 + q2)/(4πε_o(r))

Thus,

V_out = [(2.5 x 10^(-6))] + (2.7 x 10^(-6))] /(4πx8.84 x 10^(-12)x(4.8)) = 9752.14 N/C

E) V at r = 1.40 m

Corresponds with condition for eq(6)

V_out= (q1 + q2)/(4πε_o(r))

Thus,

V_out = [(2.5 x 10^(-6))] + (2.7 x 10^(-6))] /(4πx8.84 x 10^(-12)x(1.4)) 33,435.91 N/C

F) r = 0.710 Corresponds with condition for eq(5)

Thus,

V_bet = [1/(4πx8.84 x 10^(-12))]•[(2.5 x 10^(-6)/0.71) + (2.7 x 10^(-6)/1.4)] = 97318.13 N/C = 49,058 N/C

G) V at r = 0.520 m corresponds with condition for eq(5)

Thus,

V_bet = [1/(4π x 8.84 x 10^(-12))]•[(2.5 x 10^(-6)/0.52) + (2.7 x 10^(-6)/1.4)] = 60,639.65 N/C

H) V at r = 0.210 m corresponds with condition for eq(4)

V_in = [1/4πε_o]•[(q1/R1) + (q2/R2)] - - - - - (eq4)

Thus,

V_in = [1/(4π x 8.84 x 10^(-12))]•[(2.5 x 10^(-6)/0.52) + (2.7 x 10^(-6)/1.4)]

= 60,639.65 N/C

I) V at r = 0 m corresponds with condition for eq(4)

V_in = [1/4πε_o]•[(q1/R1) + (q2/R2)] - - - - - (eq4)

Thus,

V_in = [1/(4π x 8.84 x 10^(-12))]•[(2.5 x 10^(-6)/0.52) + (2.7 x 10^(-6)/1.4)]

= 60,639.65 N/C