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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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