A point charge +106.2 𝜇𝐶 is a distance 5 cm directly above the center of a square of side 10 cm,as shown in fig. What is the magnitude of the electric flux through the square? Given∈0 = 8.85 × 10−12𝑁 −1𝐶2𝑚−2
Question
A point charge +106.2 𝜇𝐶 is a distance 5 cm directly above the center of a square of side 10 cm,as shown in fig. What is the magnitude of the electric flux through the square? Given∈0 = 8.85 × 10−12𝑁 −1𝐶2𝑚−2
Solution
The electric flux (Φ) through a surface is given by Gauss's Law, which states that the electric flux through any closed surface is equal to the total charge enclosed by the surface divided by the permittivity of free space (ε0).
The formula for Gauss's Law is:
Φ = Q / ε0
Given in the problem, we have:
Q = +106.2 μC = 106.2 x 10^-6 C (converting microcoulombs to coulombs) ε0 = 8.85 x 10^-12 N^-1 C^2 m^-2
Substituting these values into Gauss's Law gives:
Φ = (106.2 x 10^-6 C) / (8.85 x 10^-12 N^-1 C^2 m^-2)
Solving this equation gives:
Φ = 1.2 x 10^16 N m^2/C
So, the magnitude of the electric flux through the square is 1.2 x 10^16 N m^2/C.
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