Question

Two spherical shells have a common center. The inner shell has radius R1 = 5.00cm and charge q1=+5.00×10^−6C ; the outer shell has radius R2 = 15.0cm and charge q2=−6.00×10^−6C . Both charges are spread uniformly over the shell surface. Take V=0 at a large distance from the shells.

Part A) What is the electric potential due to the two shells at the distance r =2.50 c from their common center.
part B) What is the electric potential due to the two shells at the distance r= 10.0 cm from their common center.

Answer 1

Step-by-step explanation:

Part A:

Using the formula for electric potential due to a charged spherical shell:

V1 = kq1/R1 and V2 = kq2/R2

where k is the Coulomb constant.

For r < R1, the electric potential due to both shells is:

V = V1 + V2 = kq1/R1 + kq2/R2

For R1 < r < R2, the electric potential due to the outer shell is:

V = V2 = kq2/R2

For r > R2, the electric potential due to both shells is:

V = V1 + V2 = kq1/r + kq2/R2

Substituting the given values, we get:

V = (9.0x10^9 Nm^2/C^2)(5.00x10^-6 C)/0.025 m + (9.0x10^9 Nm^2/C^2)(-6.00x10^-6 C)/0.15 m

V = -4.32x10^5 V

Therefore, the electric potential due to the two shells at a distance of 2.50 cm from their common center is -4.32x10^5 V.

Part B:

For r < R1, the electric potential due to both shells is:

V = V1 + V2 = kq1/R1 + kq2/R2

For R1 < r < R2, the electric potential due to both shells is:

V = V1 + V2 = kq1/r + kq2/R2

For r > R2, the electric potential due to both shells is:

V = V1 + V2 = kq1/r + kq2/r

Substituting the given values, we get:

V = (9.0x10^9 Nm^2/C^2)(5.00x10^-6 C)/0.1 m + (9.0x10^9 Nm^2/C^2)(-6.00x10^-6 C)/0.1 m

V = 4.80x10^5 V

Therefore, the electric potential due to the two shells at a distance of 10.0 cm from their common center is 4.80x10^5 V.