What is the electric field strength at points 5. 10. and 20 cm from the center?

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

General guidance

Concepts and reason

The concepts used to solve this problem are charge, charge density, and electric field.

Initially, use the relation between charge, charge density and volume to find uniform charge density of ball.

Then use the relation between charge, charge density and volume to find the amount of charge is enclosed by the spheres of radii , , and 20cm.

Finally, use the relation between electric field and distance to find the electric field strength at points , , and 20cm from the center.

Fundamentals

Volume charge density can be defined as the charge per unit volume.

The expression for the charge density is,

Here, is the charge density, is the charge, and is the volume.

Expression for the volume of the sphere is,

Here, is the radius of the sphere.

Expression for the electric field is,

Here, is the electric field, is the coulomb’s constant, and is the distance.

Step-by-step

Step 1 of 3

(a)

The expression for the charge density is,

…… (1)

Expression for the volume of the sphere is,

…… (2)

Substitute (4/3) ar? for in equation (1).

= (ਤੀਸ਼

Substitute for , for , and 20cm for .

p=
3(710)(100)
43.14)(20cm) (10m)
= 0.00212x10-C/m?
= 2.1x10C/m?

Therefore, the uniform charge density of ball is 2.1x10C/m.

Part a

The uniform charge density of ball is 2.1x10“C/m.


The charge density of ball is directly corresponds to the charge.

The charge density of ball is inversely proportional to the cube of radius of the ball.

The volume of the ball is equal to the volume of the sphere.

Step 2 of 3

(b)

The expression for the charge density is,

…… (1)

Expression for the volume of the sphere is,

…… (2)

Substitute (4/3) ar? for in equation (1).

= (ਤੀਸ਼

Rearrange the above expression for .

9
par
3 …… (3)

For the case of sphere of radius ,

Substitute 2.12x10°C/m for , for , and for in equation (3).

4(2.12x10*Сm»)(3.14)|(5cm)( 10mm)
q=-
= 1.099x10°C
1.1nC

For the case of sphere of radius ,

Substitute 2.12x10°C/mfor , for , and for in equation (3).

4(2.12x10°C°m)(3.14)|(10cm)( 10mm
q=-
= 8.87x10°C
8.9 nc

For the case of sphere of radius 20cm,

Substitute for , for , and 20cm for in equation (3).

4(2.12x10°Cm)(3.14)|(20cm) 100m)
q=-
= 71.0x10°C
710C

Therefore, theamount of charge is enclosed by the spheres of radii , , and 20cm are 1.1nC, 8.9 nc, and respectively.

Part b

The amount of charge is enclosed by the spheres of radii , 10 cm, and 20 cm are 1.1nC, 8.9nC, and 71 nC respectively.


The charge is directly proportional to the radius of the sphere.

The increase in the radius of sphere will increase the charge is enclosed by the spheres.

The sphere with radius 20cm has highest charge than other spheres.

Step 3 of 3

(c)

Expression for the electric field is,

…… (1)

For the case of the point at .

Substitute 9x10°N-m²-C2for , 1.11x10 °C for , and for in equation (1).

For the case of the point at .

Substitute 9x10°N-m²-C2for , 8.87x10°C for , and for in equation (1).

(9x10° N-m²-C2)(8.87x10°c)
((10cm)10-cm)
= 7.98x10 N/C
8.0x10 N/C

For the case of the point at 20cm.

Substitute 9x10°N-m²-C2for , 71x10°C for , and 20cm for in equation (1).

_(9x10° N-m’-c?)(71x10°C)
((20cm)(10 cm)
=1.59x10* N/C
1.6x10* N/C

Therefore, the electric field strength from the center at points , , and 20cm are 4.0x10 N/C, 8.0x10 N/C, and 1.6x10* N/C respectively.

Part c

The electric field strength from the center at points , 10 cm, and 20 cm are 4.0x10NIC, 8.0x10NIC, and 1.6x10* N/C respectively.


The electric field is directly proportional to the charge in the sphere and inversely proportional to the square of the distance.

The increase in the charge in the sphere will increase the electric field.

The increases in the distance will decrease the electric field.

Answer

Part a

The uniform charge density of ball is 2.1x10“C/m.

Part b

The amount of charge is enclosed by the spheres of radii , 10 cm, and 20 cm are 1.1nC, 8.9nC, and 71 nC respectively.

Part c

The electric field strength from the center at points , 10 cm, and 20 cm are 4.0x10NIC, 8.0x10NIC, and 1.6x10* N/C respectively.

Hottest videos

Leave a Reply

Your email address will not be published.

Related Posts