What will be the electric flux (in SI units) through a cube of side 4 cm, if a charge of 24 μC is placed at one of its corners?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 3 × 10−6 ε0−1
A charge at a corner is shared by eight identical cubes that meet at that corner. Together they form a closed surface enclosing the whole charge, so by Gauss’s law the total flux \( q/\varepsilon_0 \) divides equally among the eight cubes.
\( \displaystyle \phi = \frac{1}{8}\cdot\frac{q}{\varepsilon_0} \)
\( \displaystyle \phi = \frac{24\times10^{-6}\,\mathrm{C}}{8\,\varepsilon_0} \)
\( \displaystyle \phi = 3\times10^{-6}\,\varepsilon_0^{-1} \)
The side of the cube (4 cm) does not affect the flux.
Note: NTA picked this question from an NCERT Exemplar problem and changed the figures. Do not ignore the Exemplar while practising numericals.
NCERT Exemplar:
Electric Charges and Fields – …
