A 2.0cm radius metal sphere carries 75 nC and is surrounded by a concentric spherical conducting shell of radius 10 cm carrying -75 nC. Find the potential difference between shell and sphere.
The correct answer and explanation is:
Okay, here is the potential difference calculation and explanation.
Correct Answer:
The potential difference between the shell and the sphere is -27,000 Volts.
Explanation:
The system consists of a 2.0 cm radius conducting sphere carrying +75 nC and a concentric 10 cm radius conducting shell carrying -75 nC. Since the inner sphere is a conductor, its charge +75 nC resides uniformly on its surface. By electrostatic induction, the inner surface of the surrounding conducting shell (at radius 10 cm) must accumulate a charge of -75 nC. Because the total charge on the shell is -75 nC, the outer surface of the shell carries a net charge of (-75 nC) – (-75 nC) = 0 nC.
To find the potential difference between the shell and the sphere, we need to determine the electric potential at the surface of the shell (radius 10 cm) and the surface of the sphere (radius 2 cm). The potential difference V_shell – V_sphere is given by the negative integral of the electric field E from the sphere’s surface (r₁) to the shell’s surface (r₂):
V₂ – V₁ = V_shell – V_sphere = – ∫₁² E ⋅ dr
Using Gauss’s Law, the electric field in the region between the sphere and the shell (for r₁ < r < r₂) is due only to the charge Q₁ on the inner sphere. A spherical Gaussian surface of radius r (where r₁ < r < r₂) encloses charge Q₁. Thus, E(r) * 4πr² = Q₁/ε₀, which gives E(r) = Q₁ / (4πε₀r²) = kQ₁/r². The electric field is radial and points outward.
Now we integrate from r₁ = 0.02 m to r₂ = 0.10 m:
V_shell – V_sphere = – ∫₀.₀₂⁰.₁₀ (kQ₁ / r²) dr
= -kQ₁ ∫₀.₀₂⁰.₁₀ r⁻² dr
= -kQ₁ [-r⁻¹]₀.₀₂⁰.₁₀
= -kQ₁ [(-1/0.10) – (-1/0.02)]
= -kQ₁ [-10 – (-50)]
= -kQ₁ [40]
Using k = 9 × 10⁹ N⋅m²/C² and Q₁ = 75 × 10⁻⁹ C:
V_shell – V_sphere = -(9 × 10⁹)(75 × 10⁻⁹)(40) V
= -(9 × 75)(40) V
= -(675)(40) V
= -27000 V
The potential of the shell is 27,000 Volts lower than the potential of the sphere. This makes sense because the electric field points from the positive sphere towards the negative shell, and potential decreases in the direction of the electric field.