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

the average speed is 24.57 km/hr

Answer from: Quest

a straight line sloped

Another question on Physics

Physics, 22.06.2019 00:30

Using the work energy theorem. what is the velocity of a 850kg car after starting at rest when 13,000j of work is done to it.

Answers: 1

Answer

Physics, 22.06.2019 10:00

Which is the correct symbol for an isotope of iodine with 53 protons and 78 neutrons?

Answers: 2

Answer

Physics, 23.06.2019 06:00

Gravity is pulling an object into the sun. the closer the object gets to the sun, the greater the gravitational force that acts on the object, increasing its acceleration. a line graph plotting the speed of the object against time would have which shape? a: straight line of positive slope b: curved line that slopes upward c: curved line that slopes downward d: straight line of negative slope

Answers: 3

Answer

Physics, 23.06.2019 06:30

The energy of a photon is equal to (what) times (what) of the photon.

Answers: 1

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