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Physics Question on Electric Charges and Coulomb's Law

A point charge +10µC+10 µC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)

Answer

The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’s theorem for a cube, total electric flux is through all its six faces.
ΦTotal=qε0Φ_{Total }= \frac{q}{ε_0}
Hence, electric flux through one face of the cube i.e., through the square is
Φ=ΦTotal6=16.qε0Φ= \frac{Φ_{Total}}{6} = \frac{1}{6}.\frac{q}{ε_0}
Where,
ε0ε_0 = Permittivity of free space =8.854×10−12N−1C2m−2= 8.854 × 10^{−12} N^{−1}C^2 m^{−2}
q=10µC=10×10−6Cq = 10 µC = 10 × 10^{−6} C
Φ=16.10×10−668.854×10−12Φ = \frac{1}{6}.\frac{10 × 10^{-6}}{68.854 × 10^{-12}}
=1.88×105Nm2C−1= 1.88 × 10^5 N m^2 C^{−1}
Therefore, electric flux through the square is =1.88×105Nm2C−1.= 1.88 × 10^5 N m^2 C^{−1}.