A spherically symmetric charge distribution consists of a thick spherical shell

Question

A spherically symmetric charge distribution consists of a thick spherical shell of plastic with a uniform volume charge density of +3.00 nC/m^3 and inner and outer radii 10.0 cm and 20.0 cm; and a surrounding concentric thick spherical shell of metal with inner and outer radii 30.0 cm and 40.0 cm and excess charge -0.100 nC. Determine the magnitude and direction of the electric field at each of the following distances from the center: (a) 5.00 cm (b) 15.0 cm (c) 25.0 cm (d) 35.0 cm (e) 45.0 cm (f) What is the excess charge on the inner surface of the metal shell? (g) What is the excess charge on the outer surface of the metal shell?

Explanation / Answer

(a)

Volume of the plastic sphere is,

V = (4/3)(3.14)((0.2m)3 - (0.1 m)3) = 2.93X10-2 m3

The charge density for the sphere is = 3nC/m3

So, total charge on the sphere is,

Q = V = ( 2.93X10-2 m3)(3X10-9C/m3)

or, Q = 8.79X10-11C

At a distance 5 cm from the center the field is;

E = 0, as the point is within both the shells and field inside a shell is zero.

***************************************************************************************************

(b)

Field at distance 15 cm from the center is, say, E.

Now charge enclosed in a radius of 15 cm is, say, Q0

then Q0 = V = (4/3)(3.14)((0.15m)3 - (0.1 m)3)(3X10-9C/m3) , where V is the volume of shell of inner and outer radii, 10 cm and 15 cm respectively,

or, Q0 = 2.98X10-11C

Now we draw a spherical Gaussian surface of radius 15 cm and let the field at this point be E, then as per Gauss's law

E.4(15 cm)2 = Q0/0, where Q0 is the charge enclosed in this Gaussian surface.

So, E = (1/40)Q0/(15 cm)2 = (9X109N-m2/C2)(2.98X10-11C)/(0.15 m)2

or, E = 11.9 V/m, radially outward

***************************************************************************************************

(c)

Distance 25 cm from the center is completely outside the sphere. For points outside the sphere the charge on the sphere can be thought to be concentrated on the center of the sphere. Also the field due to metallic shell will be zero at this distance as this point falls within the metal shell. So only field will be due to the plastic sphere.

So the field is,

E = (1/40)Q/(25 cm)2, where Q = 8.79X10-11C, which we found in part (a)

or, E = (9X109N-m2/C2)(8.79X10-11C)/(0.25 m)2

or, E = 12.66 V/m, radially outward.

***************************************************************************************************

(d)

Distance 35 cm from the center falls within the metal shell itself and we know that field inside a metal is zero.

So E = 0,

***************************************************************************************************
This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification or correction, feel free to ask.....

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.