Question

A 20 cm-radius ball is uniformly charged to 71 nC.

(a) What is the ball's uniform charge density C/m^3?
(b) How much charge is enclosed by spheres of radii 5, 10, and 20 cm?
(c) What is the electric field strength at points 5, 10, and 20 cm from the center?

Answer 1

Answer:

Part a)

$\rho = 2.12\mu C/m^3$

Part b)

$q_1 = 1.11 nC$

$q_2 = 8.88 nC$

$q_3 = 71 nC$

Part c)

$E_1 = 3996 N/C$

$E_2 = 7992 N/C$

$E_3 = 15975 N/C$

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi(0.20)^3$

$V = 0.0335 m^3$

now the charge density of the ball is given as

$\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3$

Part b)

Now the charge enclosed by the surface is given as

$q = \rho V$

at radius of 5 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3$

$q = 1.11 nC$

at radius of 10 cm

$q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3$

$q = 8.88 nC$

at radius of 20 cm

$q = 71 nC$

Part c)

As we know that electric field is given as

$E = \frac{kq}{r^2}$

so we have electric field at r = 5 cm

$E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}$

$E_1 = 3996 N/C$

electric field at r = 10 cm

$E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}$

$E_2 = 7992 N/C$

electric field at r = 20 cm

$E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}$

$E_3 = 15975 N/C$

Answer 2

A) The balls uniform charge density is; 2.1188 × 10^(-6) C/m³

B) The amount of charge is enclosed by the spheres of radii 5, 10, and 20 cm are respectively;

1.1 nC, 8.875 nC, 71 nC.

C) The electric field strength at points 5, 10, and 20 cm from the center are respectively;

3960 N/C, 7987.5 N/C, 15976 N/C

Electric Field strength

A) The formula for the charge density is;

ρ = q/V

Where;

• q is magnitude of charge
• V is volume of sphere = (4/3)πr³

We are given;

• Radius; r = 20 cm = 0.2 m
• Magnitude of charge; q = 71 nC = 71 × 10^(9) C

V = (4/3) × π × 0.2³

V = 0.03351 m³

Thus;

ρ = (71 × 10^(-9))/0.03351

ρ = 2.1188 × 10^(-6) C/m³

B) Let us make q the subject in the charge density equation to get;

q = ρV

I) At r = 5cm = 0.05 m

V = (4/3) × π × 0.05³

V = 0.0005236 m³

q = 2.1188 × 10^(-6) × 0.0005236

q = 1.1 × 10^(-9) C = 1.1 nC

II) At r = 10 cm = 0.1 m

V = (4/3) × π × 0.01³

V = 0.0041888 m³

q = 2.1188 × 10^(-6) × 0.0041888

q = 8.875 × 10^(-9) = 8.875 nC

III) At r = 20 cm, as seen in the initial question,

q = 71 nC

C) The formula for the Electric field strength is;

E = kq/r²

Where k is coulombs constant = 9 × 10^(9) N.m²/C

I) At r = 5cm = 0.05 m;

E = (9 × 10^(9) × 1.1 × 10^(-9))/0.05^(2)

E = 3960 N/C

II) At r = 10 cm = 0.1 m

E = (9 × 10^(9) × 8.875 × 10^(-9))/0.1^(2)

E = 7987.5 N/C

III) At r = 20 cm = 0.2 m

E = (9 × 10^(9) × 71 × 10^(-9))/0.2^(2)

E = 15976 N/C

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