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

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Answers

answered: 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

answered: 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

answered: Guest

freezing

deposition

condensation

explanation;

melting also called fusion is the process where solid becomes liquid, sublimation is the process where solid becomes gaseous without first changing to liquid, deposition involves change from gaseous to solid state, freezing involves change from liquid to solid, boiling or evaporation involves change of liquid to gaseous state, while condensation is the change from gaseous state to liquid state.

endothermic reactions involves absorption of heat from the surroundings or the environment while exothermic reactions involves loss of heat energy to the surroundings.

in this case, sublimation, evaporation and melting are endothermic reactions (gain energy from the environment) while, freezing, deposition and condensation are exothermic (loss of energy to the surroundings) reactions.

answered: Guest

i think the answer would be reserve.