Applying Gauss’ Law : Spherical Symmetry Problems and Solutions

  Problem #1

Figure 01, gives the magnitude of the electric field inside and outside a sphere with a positive charge distributed uniformly throughout its volume. The scale of the vertical axis is set by E_s = 5.0 x 10⁷ N/C. What is the charge on the sphere?

Fig.01

Answer:
the magnitude of the electric field inside and outside a sphere with a positive charge distributed uniformly throughout its volume given by is

if r ≤ R, E = (kq/R³)r
if r > R, E = kq/r²

With r is the distance from the center of the ball.

From the graph we know that when r = R =  2 cm, the electric field is 5.0 x 10⁷ N/C, then

E = (kq/R³)r = kq/R²

5.0 x 10⁷ N/C = (9 x 10⁹ Nm²/C²)(q)/(0.02 m)

q = 2.22 x 10^─6 C = 2.22 μC

Problem #2
Two charged concentric spherical shells have radius 10.0 cm and 15.0 cm. The charge on the inner shell is 4.00 x 10^─8 C, and that on the outer shell is 2.00 x 10^─8  C. Find the electric field (a) at r = 12.0 cm and (b) at r = 20.0 cm.

Answer:
Known:
concentric spherical radius r₁ = 10.0 cm and r₂ = 15.0 cm
The charge on the inner shell is q₁ = 4.00 x 10^─8 C,
and that on the outer shell is q₂ = 2.00 x 10^─8  C

(a) the electric field at r = 12.0 cm; [r₁ = 10.0 cm < r = 12.0 cm < r₂ = 15.0 cm]

E(r) = kq₁/r²
        =  (9 x 10⁹ Nm²/C²)(4.00 x 10^─8 C)/(0.20 m)²
E(r) = 2.5 x 10⁴ N/C

(b) the electric field at r = 20.0 cm or r₁ <  r₂ < r

E(r) = k(q₁ + q₂) /r²
        =  (9 x 10⁹ Nm²/C²)(4.00 x 10^─8 C + 2.00 x 10^─8  C)/(0.20 m)²
E(r) = 1.35 x 10⁴ N/C

Problem #3
An unknown charge sits on a conducting solid sphere of radius 10 cm. If the electric field 15 cm from the center of the
sphere has the magnitude 3.0 x 10³ N/C and is directed radially inward, what is the net charge on the sphere?

Answer:
Known:
Radius conducting solid sphere, R = 10 cm = 0.1 m
Electric field, E = 3.0 x 10³ N/C
As
E = kq/R²

3.0 x 10³ N/C  = (9 x 10⁹ Nm²/C²)(q)/(0.15 m)²

q = 7.5 x 10^─9 C = 7.5 nC

Here, since electric field is directed radially inward charge q is ngative

Thus, q = ─7.5 x 10^─9 C = ─7.5 nC

Problem #4
A charged particle is held at the center of a spherical shell. Figure 02 gives the magnitude E of the electric field versus radial distance r. The scale of the vertical axis is set by E_s = 10.0 x 10⁷ N/C. Approximately, what is the net  charge on the shell?

Fig.02

Answer:
the magnitude of the electric field inside and outside a sphere with a positive charge distributed uniformly throughout its volume given by is

if r ≤ R, E = (kq/R³)r
if r > R, E = kq/r²

for if r ≤ R = 2.5 cm = 2.50 x 10^─2 m and E = 2.0 x 10⁷ N/C (from graph), then

E = (kq/R³)R = kq/R²

2.0 x 10⁷ N/C = (9 x 10⁹ Nm²/C²)(q)/(0.025 m)²

q = 1.4 x 10^─6 C

for if r > R, r = 3.0 cm = 3.0 x 10^─2 m and E = 8.0 x 10⁷ N/C (from graph), then

E = (kq/R³)R = kq/R²

8.0 x 10⁷ N/C = (9 x 10⁹ Nm²/C²)(Q)/(0.03 m)²

Q = 8.0 x 10^─6 C

Net change on the shell: Q_net = Q ─ q = 6.6 x 10^─6 C

Problem #5
In Fig. 03, a solid sphere of radius a = 2.00 cm is concentric with a spherical conducting shell of inner radius b = 2.00a and outer radius c = 2.40a. The sphere has a net uniform charge q₁ = 5.00 fC; the shell has a net charge q₂ = ─q₁. What is the magnitude of the electric field at radial distances (a) r = 0, (b) r = a/2.00, (c) r = a, (d) r = 1.50a, (e) r = 2.30a, and (f) r = 3.50a? What is the net charge on the (g) inner and (h) outer surface of the shell?

Fig.03

Answer:
Known:
net uniform charge  q₁ = 5.00 fC =  5.00 x 10^-15 C
net charge q₂ = ─q₁ = ─ 5.00 x 10^-15 C
of radius solid sphere is a = 2.00 cm = 2.00 x 10^-2 m

the magnitude of the electric field inside a sphere with a positive charge distributed uniformly throughout its volume given by is

if r ≤ a, E = (kq/a³)r

then,
(a) the magnitude of the electric field at radial distances r = 0 is,

E = (kq/a³)r = 0

(b) the magnitude of the electric field at radial distances r = a/2.00 is,

E = (kq/a³)(a/2.00)

E = (9 x 10⁹ Nm²/C²)(5.00 x 10^-15 C)(1/2)/(2.00 x 10^-2 m)²

E = 5.62 x 10^-2 N/C

(c) the magnitude of the electric field at radial distances r = a is,

 E = (kq/a³)(a) = kq/a²

E = (9 x 10⁹ Nm²/C²)(5.00 x 10^-15 C)/(2.00 x 10^-2 m)²

E = 0.112 N/C

(d) the magnitude of the electric field at radial distances r = 1.50a is,

the magnitude of the electric field outside a sphere with a positive charge distributed uniformly throughout its volume given by is

if r ≥ a, E = kq/r²

then,

E = kq/r² = kq/(1.5a)²

E = (9 x 10⁹ Nm²/C²)(5.00 x 10^-15 C)/(1.5 x 2.00 x 10^-2 m)²

E = 0.05 N/C

(e) the magnitude of the electric field at radial distances r = 2.30a, when a < r < c, then E = 0.

(f) the magnitude of the electric field at radial distances r = 3.50a,

∑q = q₁ + q₂ = q₁ + (-q₁) = 0, then

E = kq/r² = 0

(g) the net charge on the inner,

Electric field, E = 0 N/C

and electric flux, φ = 0

because, φ = q_encl/ϵ₀, then q_encl = 0

so that

q_spere + q_inner = 0

q_inner = –q_spere = –5.00 x 10^-15 C

(h) the net charge on the inner,

Fig.04

(h) the net charge on the outer surface of the shell

ΣQ_shell = –5 fC

Q_i + Q_o = –5 fC

–5 fC + Q_o = –5 fC

Q_o = 0

Problem #6
Figure 05 shows, in cross section, two solid spheres with uniformly distributed charge throughout their volumes. Each has radius R. Point P lies on a line connecting the centers of the spheres, at radial distance R/2.00 from the center of sphere 1. If the net electric field at point P is zero, what is the ratio q₂/q₁ of the total charges?
  

Fig.05

Answer:
For q₁, the magnitude of the electric field inside a sphere with a positive charge distributed uniformly throughout its volume given by is

if r ≤ R, E = (kq/R³)r
Then,

E₁ = (kq₁/R³)r₁

E₁ = (kq₁/R³)(R/2) = kq₁/2R²

For q₂, the magnitude of the electric field outside a sphere with a positive charge distributed uniformly throughout its volume given by is

if r ≥ R, E = kq/r²

Then,

E = kq₂/r², with r = 1.50 R,

E = kq/(1.50R)² = 4kq₂/9R²

If the net electric field at point P is zero, then

E₁ = E₂

kq₁/2R² = 4kq₂/9R²

so, q₂/q₁ = 9/8 = 1.125  

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