Question

Question in attachment
​

Answer 1

$\huge{\underline{\underline{\bf{Answer:-}}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Electric flux and Gauss law:

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

The electric flux is defined as the dot product of the electric field and the area vector.

⠀⠀⠀⠀⠀⠀⠀⠀⠀

• According to Gauss law, the electric flux is given as follows:

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$ϕ= \frac{Q}{ϵo}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Where,

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• Q is the charge enclosed in the gaussian surface.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

• ϵo is the electric permittivity of the free space.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

${\purple{\underline{\underline{\mathsf{\pink{Answer\:and\:Explanation:}}}}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Given Data:

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$charge \: 1,(q1) = 3.4 \: nC$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$charge \: 2,(q2) = - 6.00 \: nC$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

(A)

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Now, the charge 2 will be inside the sphere of the radius $( {r}^{2} ) = 1.55 \: m,$therefore the enclosed charge would be

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$Q = q_{2}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$Q = - 6 \: nC$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Therefore, the electric flux would be

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$ϕ= \frac{Q}{ϵo}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$= \frac{ - 6 \times 10 {}^{ - 2}C }{8.8 \times 10 {}^{ - 12} }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$= - 678 \: N {m}^{2} C$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

(B)

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

In this sphere both charge will be inside the sphere.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$Q = q_{1} + q_{2}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$Q = 3.4 - 6$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$Q = - 2.6nC$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Therefore, the electric flux would be

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$ϕ= \frac{Q}{ϵo}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$= \frac{ - 2.6 \times 10 {}^{ - 2}C }{8.5 \: \times 10 {}^{ - 12} }$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$= - 263 \: Nm {}^{2} C$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

Answer 2

Answer:

The answer is

Electric flux and Gauss law.

Pls mark me the brainliest!!