Answer:
a) -2.516 × 10⁻⁴ V
b) -1.33 × 10⁻³ V
Explanation:
The electric field inside the sphere can be expressed as:
[tex]E= \frac{kqr}{R^3}[/tex]
The potential at a distance can be represented as:
V(r) - V(0) = [tex]-\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2[/tex]
V(r) - V(0) = [tex][\frac{qr^2}{8 \pi E_0R^3 }][/tex]₀
V(r) = [tex]-[\frac{qr^2}{8 \pi E_0R^3 }][/tex]₀
Given that:
q = +3.83 fc = 3.83 × 10⁻¹⁵ C
r = 0.56 cm
= 0.56 × 10⁻² m
R = 1.29 cm
= 1.29 × 10⁻² m
E₀ = 8.85 × 10⁻¹² F/m
Substituting our values; we have:
[tex]V(r)[/tex] [tex]= -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}[/tex]
[tex]V(r)[/tex] = -2.15 × 10⁻⁴ V
The difference between the radial distance and center can be expressed as:
V(r) - V(0) = [tex]-\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2[/tex]
V(r) - V(0) = [tex][\frac{qr^2}{8 \pi E_0R^3 }]^R[/tex]
V(r) = [tex]-\frac{qR^2}{8 \pi E_0R^3 }[/tex]
V(r) = [tex]-\frac{q}{8 \pi E_0R }[/tex]
V(r) [tex]= -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}[/tex]
V(r) = -0.00133
V(r) = - 1.33 × 10⁻³ V
