A flat, square surface with sides of length \(L\) is described by the equations \(x=L, 0 \leq y \leq L, 0…
\(x=L, 0 \leq y \leq L, 0 \leq z \leq L\)
Find the electric flux through the square due to a positive point charge \(q\) located at the origin \((x=0, y=0, z=0)\)
- \(\frac{q}{4 \varepsilon_{0}}\)
- \(\frac{q}{6 \varepsilon_{0}}\)
- \(\frac{q}{24 \varepsilon_{0}}\)
- \(\frac{q}{48 \varepsilon_{0}}\)
Solution
Imaging a charge \(q\) at the center of a cube of edge length \(2 \mathrm{~L}\) (figure). Then:
\(\phi=\frac{q}{\varepsilon_{0}}\)
Here, the square is one \(24^{\text {th }}\) of the surface area of the imaginary cube, so it intercepts \(1 / 24\) of the flux. That is,
\(\phi=\frac{q}{24 \varepsilon_{0}}\) ^
Asked in: JEE Mains - Electrostatics - Chapter Test