Question

A point charge of 10 to the power minus 7 coulomb is situated at the centre of a cube of side 1 m find electric flux through its surface.

Answer 1

Question 20
A point charge causes an electric flux of −1.0 × 103Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge.(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?(b) What is the value of the point charge?Answer

(a) Electric flux, Φ = −1.0 × 103 N m2/C

Radius of the Gaussian surface,

r = 10.0 cm

Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., −103N m2/C.

(b) Electric flux is given by the relation,

Where, q = Net charge enclosed by the spherical surface

∈0 = Permittivity of free space = 8.854 × 10−12 N−1C2 m−2

∴ 

= −1.0 × 103 × 8.854 × 10−12

= −8.854 × 10−9 C

= −8.854 nC

Therefore, the value of the point charge is −8.854 nC.

Answer 2

The electric flux through its surface is $1.12\times 10^{4}\ Nm^2/C$.

Explanation:

Given that,

Charge, $q=10^{-7}\ C$

Side of the cube, a = 1 m

To find,

Electric flux through its surface.

The electric flux through its surface is given by Gauss's law of electrostatics. The electric flux is given by :

$\phi=\dfrac{q}{\epsilon_o}$

$\phi=\dfrac{10^{-7}}{8.85\times 10^{-12}}$

$\phi=1.12\times 10^{4}\ Nm^2/C$

So, the electric flux through its surface is $1.12\times 10^{4}\ Nm^2/C$. Hence, this is the required solution.

Learn more,

Gauss's law of electrostatics

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