Question

A square of side 20 cm is enclosed by a surface
of sphere of 80 cm radius. Square and sphere have
the same centre. Four charges. + 4 x 10-° C,
-5 x 10-6C, -3 x 10-6C, + 6 x 10C are located
at the four corners of a square, then total flux from
spherical surface in N-m?/C Will be :
(1) zero
(2) 481 x 103
(3) 721 x 103 (4) 361 x 103​

Answer 1

Answer:

Total flux passing through the spherical surface must be ZERO

Explanation:

As we know that total flux passing through a closed surface is given by ratio of total enclosed charge and permitivity of the space

so we have

$\phi = \frac{Q}{\epsilon_0}$

now we have

$Q = q_1 + q_2 + q_3 + q_4$

$Q = 2 \times 10^{-6} - 5 \times 10^{-6} - 3 \times 10^{-6} + 6 \times 10^{-6}$

$Q = ZERO$

now we have

$\phi = \frac{0}{8.85 \times 10^{-12}}$

$\phi = 0$

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Topic : electric Flux

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Answer 2

Answer:

Total flux passing through the spherical surface must be 72pi X 10^3

Explanation:

Explanation:

As we know that total flux passing through a closed surface is given by ratio of total enclosed charge and permittivity of the free space

so we have ,

Ф = ∑Q/∈₀

Ф=Flux , ∑Q= summation of all charges, ∈₀= permitivity of space

now we have ,

∑Q= +4X10^-6 +(-5X10^-6)+(-3X10^-6)+6X10^-6

     = +2X10^-6

Then,

Total Fiux  (Ф) = (+2X10^-6)/∈₀

(SInce the value of ∈₀ is 8.85X10^-12 But we write it as 1/(4πX9X10^9))        

Thus we have , Ф= +2X10^-6X4πX9X10^9

       (The denominator will become numerator when it will be divided )

So, Ф= 72πX10^3

Hence , Option (3) is Correct.