A conducting sphere with a diameter of $4.00cm$ is charged so that its electric potential is $50V$ relative to infinity. What is the total charge on the sphere? (Assume the sphere is isolated in air, and use $\epsilon 0_0$ for the permittivity of free space.)

• A. $2.22\times 10{\mathrm{^\{-12\}}}^{1}C$
• B. $8.85\times 10{\mathrm{^\{-12\}}}^{1}C$
• C. $2.22\times 10{\mathrm{^\{-11\}}}^{1}C$
• D. $1.11\times 10{\mathrm{^\{-11\}}}^{1}C$

Answer: (C)

Verified step by step guidance

1. Identify the given quantities: the diameter of the conducting sphere is 4.00 cm, so the radius \(r\) is half of that, which is \(r = \frac{4.00}{2} = 2.00\) cm. Convert this to meters for SI units: \(r = 2.00 \times 10^{-2}\) m. The electric potential \(V\) of the sphere relative to infinity is 50 V.
2. Recall the formula for the electric potential \(V\) on the surface of a charged conducting sphere of radius \(r\) carrying charge \(Q\) in vacuum (or air): \(V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}\) where \(\varepsilon_0\) is the permittivity of free space.
3. Rearrange the formula to solve for the total charge \(Q\) on the sphere: \(Q = 4 \pi \varepsilon_0 r V\)
4. Substitute the known values into the equation: use \(\varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2\), \(r = 2.00 \times 10^{-2}\) m, and \(V = 50\) V.
5. Calculate the product step-by-step to find the total charge \(Q\) on the sphere, ensuring units are consistent and the final charge is expressed in coulombs (C).