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Spheres of Charge: A metal sphere of radius 10 cm carries an excess charge of +2.0 μC. What is the magnitude of the electric field 5.0 cm above the sphere's surface? (k = 1/4πε0 = 9.0 × 10^9 N ∙ m^2/C^2)

Answer :

cjmejiab

To solve this problem we will apply the concept related to the electric field.  The magnitude of each electric force with which a pair of determined charges at rest interacts has a relationship directly proportional to the product of the magnitude of both, but inversely proportional to the square of the segment that exists between them. Mathematically can be expressed as,

[tex]E = \frac{kV}{r^2}[/tex]

Here,

k = Coulomb's constant

V = Voltage

r = Distance

Replacing we have

[tex]E = \frac{(9*10^9)(2*10^{-6})}{((10+5)*10^{-2})^2}[/tex]

[tex]E = 8*10^5N/C[/tex]

Therefore the magnitude of the electric field is [tex]8*10^5N/C[/tex]

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