Question

In: Physics

A ) A 2-?C point charge is embedded in the center of a solid Pyrex sphere...

A ) A 2-?C point charge is embedded in the center of a solid Pyrex sphere of radius R = 15 cm.

(1) Calculate the electric field strength E just beneath the surface of the sphere. (in N/C ) - Answer them in term of "e" like [ ex. : 1.43e+05? ]

(2) Assuming that there are no free charges, calculate the strength of the electric field just outside the surface of the sphere. ( in N/C ) Answer them in term of "e" like [ ex. : 1.43e+05? ]?

(3) What is the induced surface charge density ?ind on the surface of the Pryex? (in C/m^2) Answer them in term of "e" like [ ex. : 1.43e+05? ]?

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B) Trying with diffrent variables, a 1-?C point charge is embedded in the center of a solid Pyrex sphere of radius R = 12 cm.

(1) Calculate the electric field strength E just beneath the surface of the sphere.( in N/C )? Answer them in term of "e" like [ ex. : 1.43e+05? ]?

(2) Assuming that there are no free charges, calculate the strength of the electric field just outside the surface of the sphere.( in N/C )? Answer them in term of "e" like [ ex. : 1.43e+05? ]?

(3) What is the induced surface charge density ?ind on the surface of the Pryex?(in C/m^2) Answer them in term of "e" like [ ex. : 1.43e+05? ]

Expert Solution

Dielectric const. of pyrx glass K = 5.6

= k _o

1. Draw a Guassian surface concentric with the cneter of the surface just beneath the surface

From Gauss law

E_in* 4R² = Q/

charge enclosed inside the surface Q = 2.0C

radius of the sphere R = 0.12 m

E_in = 9.0e+9 *2.0 / 5.6 * 0.12²

= 2.23e+11 N/C

2. just out isde the sphere we draw a a Gaussian surface same as earlier but now it is in air where K=1 and

= _o

The field just outside the surface

E_out = 4/_o * Q/R²

= 9.0e+9 * 2.0 / 0.12²

= 1.25e+12 N/C

3. The differenc ein fields is due to the presence of pyrex which is a dieelectric. The dielectric is polarised and produces and causes an electric field opposite to the original field and hence the field just beneath the surface is less

The difference in electric fields

E_out - E_in = 4/_o * Q/R² ( 1-1/K)

the induced charge = Q(1-1/K) = 2.0* (1-1/5.6) = 1.64 C

genius_generous answered 1 year ago

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