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The electric field at the surface points radially outward and has a magnitude of 8300 n/c. if r1 = 0.5 m, r2 = 1.9 m, and a = 3, find the value of the charge q, in coulombs?

Sagot :

The value of charge in coulombs is 0.9 × 10—⁹ coulombs.

Radius 1 of the surface = 0.5 m

Radius 2 of the sphere = 1.9 m

Area of the surface = 3

The magnitude of the electric field at the surface = 8300 N/C

The electric flux through a surface is,

[tex] Electric \: flux=Electric \: field \: \times area \: of \: the \: surface[/tex]

[tex]Φ = E \times S[/tex]

[tex]Φ = EScosθ[/tex]

[tex]Φ =E \times 4 \times \pi \times r ^{2} [/tex]

According to Gauss, the net electric flux is,

[tex]Net \: electric \: flux = \frac{1}{E _{0} } \times charge \: enclosed[/tex]

[tex]Charge \: enclosed = 4 \times \pi \times E _0 \times a \times r ^{3} [/tex]

[tex] = \frac{1}{9 \times 10 ^{9} } \times 3 \times (1.4) ^{3} [/tex]

[tex] = 0.9 \times 10 ^{ - 9} \: C[/tex]

Therefore, the value of charge in coulombs is 0.9 × 10—⁹ coulombs.

To know more about the electric field, refer to the below link:

https://brainly.in/question/7556475

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