In a parallel plate capacitor with air between the plates, each plate has an area of \( 6\times10^{-3} \) m\( ^{2} \) and the separation between the plates is 2 mm. The charge on each plate, if the capacitor is connected to a 200 V supply, will be
(Given: \( \epsilon_{0}=8.85\times10^{-12} \) C\( ^{2} \)N\( ^{-1} \)m\( ^{-2} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \sim 5.3\times10^{-9} \) C
The charge on a capacitor is \( Q = CV \), and for a parallel plate capacitor with air between the plates \( C = \epsilon_0A/d \).
\( A = 6\times10^{-3}\,\mathrm{m^{2}} \)
\( d = 2\,\mathrm{mm} = 2\times10^{-3}\,\mathrm{m} \)
\( V = 200\,\mathrm{V} \)
\( \displaystyle C = \frac{\epsilon_0A}{d} = \frac{8.85\times10^{-12}\times6\times10^{-3}}{2\times10^{-3}} \)
\( C = 26.55\times10^{-12}\,\mathrm{F} = 2.655\times10^{-11}\,\mathrm{F} \)
\( Q = CV = 2.655\times10^{-11}\times200 = 5.31\times10^{-9}\,\mathrm{C} \)
\( Q \approx 5.3\times10^{-9}\,\mathrm{C} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
