A conducting sphere of radius $20 \mathrm{~cm}$ is charged so that the electric field reaches $3.0 \times 10^{5} \mathrm{~V} / \mathrm{m}$ at its surface. What is the potential at the surface of the sphere?

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The potential at the surface of a conducting sphere is given by:$$V = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r}$$where:* $V$ is the potential* $Q$ is the charge on the sphere* $r$ is the radius of the sphere* $\varepsilon_0$ is the permittivity of free spaceWe are given that the electric field at the surface of the sphere is $3.0 \times 10^{5} \mathrm{~V} / \mathrm{m}$. The electric field is related to the potential by:$$E = -\frac{dV}{dr}$$where:* $E$ is the electric field* $V$ is the potential* $r$ is the distance from the center of the sphereWe can use this equation to find the potential at the surface of the sphere:$$V = -\int_{r}^{\infty} E dr$$$$V = -\int_{0.2}^{0.2} 3.0 \times 10^{5} dr$$$$V = -3.0 \times 10^{5} (0.2 - 0)$$$$V = -6.0 \times 10^{4} \mathrm{~V}$$Therefore, the potential at the surface of the sphere is $-6.0 \times 10^{4} \mathrm{~V}$.

A Conducting Sphere of Radius 20 cm20 \mathrm{~cm}20 cm Is Charged So That the Electric Field Reaches

A Conducting Sphere of Radius $20 \mathrm{~cm}$ Is Charged So That the Electric Field Reaches