For this part of exam, your detailed solutions including starting assumptions or

Question

For this part of exam, your detailed solutions including starting assumptions or physical laws, physical or logical reasoning, and mathematical calculations will he examined and evaluated. They are as important as your actual results. Make sure you give clear and adequate, hut short, explanations A spherical conductor of radius 3.00 cm has a charge of -2.21 times 10-8C. This conductor is centered inside a hollow metal conducting spherical shell which extends from 5.00 to 8.00 cm. The total charge on the spherical shell is unknown. At a point 10.0 cm from the centre of the sphere and the shell, the electric field due to the charges on these two conductors is 21.0 kN/C directed radially outward. Using Gauss' Law, determine the charge on the inner surface of the shell. In similar fashion, use Gauss' Law to find the charge on the outer surface of the shell. What arc the magnitude and direction of the electric field at 2.00.4.00 and 6.00 cm from the centre of the system?

Explanation / Answer

a)

As the field inside the conductor must be Zero ,

Then , using Gauss law

(Qin - 2.21 *10^-8) = 0/epsilon

Qin = 2.21 *10^-8 C

the charge on the inner surface is 2.21 *10^-8 C

b)

Now, let Q be the charge on the outer surface ,

THen , Using gauss law .

21 *10^3 * 4 *pi*0.10^2 = Q/8.854 *10^-12

Q = 2.34 *10^-8 C

the charge on outer surface is 2.34 *10^-8 C

c)
at x = 2 cm

Electric field = 0 N/C

at x = 4 cm

E = 9*10^9 * 22.1 *10^-9/.04^2

E = 1.243 *10^5 N/C towards the centre

at x = 6 cm

inside the conductor ,

ELectric field is ZERO

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

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