The flux of electrostatic field through the closed spherical S' is found to be four times that through the closed spherical surface S. Find the magnitude of the charge Q . Given Q1 = 1μC , q2 = - 2 μC , q3 = 9.84 μC .
Answers
Answered by
16
Explanation:
Given The flux of electrostatic field through the closed spherical S' is found to be four times that through the closed spherical surface S. Find the magnitude of the charge Q Given Q1 = 1μC , q2 = - 2 μC , q3 = 9.84 μC .
- According to Gauss law total electric flux through the closed surface
- So φ = ∫ E. ds
- = q / εo
- According to the question the flux through the closed surface will be
- 4 S = q1 / εo + q2 / εo + q3 / εo
- 4 (Q / εo) = q1 / εo + q2 / εo + q3 / εo
- 4Q = q1 + q2 + q3
- 4Q = 1 x 10^-6 – 2 x 10^-6 + 9.84 x 10^-6
- 4Q = 10^-6 (1 – 2 + 9.84)
- 4Q = 8.84 x 10^-6
- Or Q = 2.21 x 10^-6 C
Reference link will be
https://brainly.in/question/17070865
Answered by
20
Answer:
Q = 26.52 μC or 2.652 x 10^-5
Explanation:
Now given,
flux of S' = 4 x flux of S
so for a closed surface,
(Q + q1 + q2 + q3 )/ε₀ = 4 (q1 + q2 + q3 )/ε₀
Cutting ε₀ on both sides we get
Q + q1 + q2 + q3 = 4 ( q1 + q2 + q3 )
Q = 4 ( q1 + q2 + q3 ) - ( q1 + q2 + q3 )
Q = ( q1 + q2 + q3 )( 4 - 1 )
Q = 3 ( q1 + q2 + q3 )
Q = 3 ( 1 - 2 + 9.84 ) = 3 ( 8.84 ) = 26.52 μC or 2.652 x 10^-5
Hope this helps!
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