Answer :
The potential difference between the center and the surface of the sphere is [tex]\(8.99 \times 10^7 \, \text{V}\)[/tex].
To find the potential difference between the center and the surface of the sphere, we can use the formula for the electric potential due to a uniformly charged sphere:
[tex]\[V = \frac{kQ}{R},\][/tex]
where (V) is the electric potential, (k) is Coulomb's constant [tex](\(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\))[/tex], (Q) is the total charge, and (R) is the radius of the sphere.
Given [tex]\(Q = +6.00 \, \mu\text{C}\) and \(R = 6.00 \, \text{cm} = 0.06 \, \text{m}\)[/tex], we can plug these values into the formula:
[tex]\[V = \frac{(8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \times (6.00 \times 10^{-6} \, \text{C})}{0.06 \, \text{m}}.\][/tex]
Calculating this expression yields:
[tex]\[V = 8.99 \times 10^7 \, \text{V}[/tex]
The potential difference between the center and the surface of the sphere is 1.08 × 10⁶ V.
What is the voltage difference between the center and surface of an insulating sphere?
The potential difference between the center and surface of the sphere with a uniform charge distribution can be expressed mathematically as:
V = kQ/R
Where V is the potential difference, k is Coulomb's constant (9 × 10⁹ Nm²/C²), Q is the charge, and R is the radius of the sphere.
For this specific problem, the potential difference can be calculated as:
V = (9 × 10⁹ Nm²/C²) × (+6.00 × [tex]10^-^6[/tex] C) / (0.06 m) = 1.08 × 10⁶ V.
Therefore, the potential difference between the center and surface of the sphere is 1.08 × 10⁶ V.
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