A charge $+q$ is at a distance $L / 2$ above a square of side $\mathrm{L}$. Then what is the flux linked…
- $\frac{\mathrm{q}}{4 \varepsilon_{0}}$
- $\frac{2 q}{3 \varepsilon_{0}}$
- $\frac{\mathrm{q}}{6 \varepsilon_{0}}$
- $\frac{6 \mathrm{q}}{\varepsilon_{0}}$
Solution
The given square of side $\mathrm{L}$ may be considered as one of the faces of a cube with edge $\mathrm{L}$. Then given charge $\mathrm{q}$ will be considered to be placed at the centre of the cube. Then according to Gauss's theorem, the magnitude of the electric flux through the faces (six) of the cube is given by $\phi=\mathrm{q} / \varepsilon_{0}$ Hence, electric flux through one face of the cube for the given square will be $\phi^{\prime}=\frac{1}{6} \phi=\frac{\mathrm{q}}{6 \varepsilon_{0}}$
Asked in: MHT CET Full Test 12