Respuesta :
Answer:
Explanation:
To solve this problem we can use the Gauss' Theorem
Hence, we have:
[tex]\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}[/tex]
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
[tex]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}[/tex]
b) r=0.70m (R1<r<R2)
QN=+2.50 μC
[tex]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}[/tex]
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
[tex]V=-\int \frac{Q_N}{4\pi r^2\epsilon_0}dr=\frac{Q_N}{4\pi r\epsilon_0}[/tex]
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
[tex]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}[/tex]
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:
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
