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A spherical conductor with a radius of 54.5 cm has an electric field of magnitude 4.00 × 105 V/m just outside its surface. What is the electric potential just outside the surface, assuming the potential is zero far away from the conductor?

Respuesta :

Answer:

 [tex] V =2.18 \times 10^5\ V[/tex]

Explanation:

given,

radius of the sphere, R = 54.5 cm

                                   R = 0.545 m

Electric magnitude of the conductor,E = 4 x 10⁵ V/m

Electric potential just outside of the surface,V = ?

we know,

 [tex]E = \dfrac{V}{r}[/tex]

 [tex] V = E.r[/tex]

 [tex] V =4\times 10^5 \times 0.545[/tex]

 [tex] V =2.18 \times 10^5\ V[/tex]

Hence, the Potential just outside the surface is equal to [tex] V =2.18 \times 10^5\ V[/tex]

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