KCET · Physics · Electrostatics
A point charge \(q\) is placed at the corner of a cube of side \(a\) as shown in the figure. What is the electric flux through the face \(A B C D\) ?
- A 0
- B \(\frac{q}{24 \varepsilon_{0}}\)
- C \(\frac{q}{6 \varepsilon_{0}}\)
- D \(\frac{q}{72 \varepsilon_{0}}\)
Answer & Solution
Correct Answer
(B) \(\frac{q}{24 \varepsilon_{0}}\)
Step-by-step Solution
Detailed explanation
If charge is placed at the one corner of a cube, then it can be shared further by 8 cubes.
\(\therefore\) Charge on one cube \(=\frac{q}{8}\)
Now, in any given cube this charge will be touching 3 out of 9 of cube faces. So, the area vector of that side and the electric field vector will be perpendicular. So, flux through those 3 sides will be zero. However, equal amount of flux will flow from other 3 sides.
So, flux through one side \(=\left(\frac{q / 8}{3}\right) \cdot \frac{1}{\varepsilon_{0}}=\frac{q}{24 \varepsilon_{0}}\) Hence, flux through one face \((A B C D)=\frac{q}{24 \varepsilon_{0}}\)
\(\therefore\) Charge on one cube \(=\frac{q}{8}\)
Now, in any given cube this charge will be touching 3 out of 9 of cube faces. So, the area vector of that side and the electric field vector will be perpendicular. So, flux through those 3 sides will be zero. However, equal amount of flux will flow from other 3 sides.
So, flux through one side \(=\left(\frac{q / 8}{3}\right) \cdot \frac{1}{\varepsilon_{0}}=\frac{q}{24 \varepsilon_{0}}\) Hence, flux through one face \((A B C D)=\frac{q}{24 \varepsilon_{0}}\)
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