Question

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 .

Answer 1

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

Answer 2

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!