Physics, 31.07.2019 18:10 aakakaa

A20 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?

Answers: 3

Show answers

Answers

Answer from: jayowens20

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 chargeV is volume of sphere = (4/3)πr³

We are given;

Radius; r = 20 cm = 0.2 mMagnitude 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

Read more on electric field strength at; link

Answer from: jazzclark79

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 from: Quest

Bsame force but in the same direction

Answer from: Quest

neither accurate nor precise.

Another question on Physics

Physics, 22.06.2019 10:30

Aperson gives a shopping cart an initial push along a horizontal floor to get it moving, and then lets go. the cart travels forward along the floor, gradually slowing as it moves. consider the horizontal force(s) on the cart while it is moving forward and slowing. which of the following statements is correct? (a) both a forward and a backward force are acting on the cart, but the forward force is larger. (b) only a forward force is acting which diminishes with time. (c) both a forward and a backward force are acting on the cart, but the backward force is larger. (d) only a backward force is acting, no forward force.

Answers: 3

Answer

Physics, 22.06.2019 11:00

If angle "a" is 25°, angle "b" is: a) equal to angle c b) less than angle c c) greater than angle c

Answers: 1

Answer

Physics, 22.06.2019 12:10

An ionic compound can conduct electricity when a it is dissolved in water b its crystals are ground into powder c it is in solid form d it is heated to just below its melting point

Answers: 1

Answer

Physics, 22.06.2019 14:20

What are the starting materials for nuclear fission? two small nuclei two large nuclei a neutron and a large nucleus a neutron and a small nucleus

Answers: 2

Answer

You know the right answer?

A20 cm-radius ball is uniformly charged to 71 nc.
(a) what is the ball's uniform charge densit...

Questions