Question:
216 identical spherical drops each having a positive charge of
\[
10\,\text{nC}
\]
combine to form a big spherical drop. If the radius of each small drop is
\[
3\,\text{mm},
\]
then the capacitance and electric potential of the big drop are respectively
216 identical spherical drops each having a positive charge of
\[
10\,\text{nC}
\]
combine to form a big spherical drop. If the radius of each small drop is
\[
3\,\text{mm},
\]
then the capacitance and electric potential of the big drop are respectively
Show Hint
When \(n\) identical drops combine:
\[
R=n^{1/3}r,
\qquad
Q=nq.
\]
For a spherical conductor:
\[
C=4\pi\varepsilon_0R,
\qquad
V=\frac{Q}{C}.
\]
The potential increases as \(n^{2/3}\).
Updated On: Jul 29, 2026
- \(20\,\text{pF},\ 1.08\times10^6\,\text{V}\)
- \(20\,\text{pF},\ 1080\,\text{V}\)
- \(2\,\text{pF},\ 1.08\times10^6\,\text{V}\)
- \(2\,\text{pF},\ 1080\,\text{V}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
For an isolated spherical conductor,
\[
C=4\pi\varepsilon_0R
\]
and
\[
V=\frac{Q}{C}.
\]
When \(n\) identical drops combine,
\[
R=n^{1/3}r.
\]
Step 1: Find the radius of the big drop. Given, \[ n=216, \qquad r=3\,\text{mm}. \] Since \[ 216=6^3, \] \[ R=216^{1/3}\times3\,\text{mm}. \] \[ R=6\times3\,\text{mm}. \] \[ R=18\,\text{mm}. \] \[ R=0.018\,\text{m}. \]
Step 2: Calculate the capacitance of the big drop. \[ C=4\pi\varepsilon_0R. \] Using \[ 4\pi\varepsilon_0=\frac1{9\times10^9}, \] \[ C=\frac{0.018}{9\times10^9}. \] \[ C=2\times10^{-12}\,\text{F}. \] \[ \boxed{C=2\,\text{pF}}. \]
Step 3: Find the total charge on the big drop. Each drop carries \[ q=10\,\text{nC}=10\times10^{-9}\,\text{C}. \] Hence, \[ Q=216q. \] \[ Q=216\times10\times10^{-9}. \] \[ Q=2.16\times10^{-6}\,\text{C}. \]
Step 4: Calculate the potential of the big drop. \[ V=\frac{Q}{C}. \] \[ V = \frac{2.16\times10^{-6}} {2\times10^{-12}}. \] \[ V = 1.08\times10^6\,\text{V}. \] Therefore, \[ \boxed{V=1.08\times10^6\,\text{V}} \] \[ \boxed{ \text{Capacitance}=2\,\text{pF}, \qquad \text{Potential}=1.08\times10^6\,\text{V} } \] \[ \boxed{\text{Answer = (C)}} \]
Step 1: Find the radius of the big drop. Given, \[ n=216, \qquad r=3\,\text{mm}. \] Since \[ 216=6^3, \] \[ R=216^{1/3}\times3\,\text{mm}. \] \[ R=6\times3\,\text{mm}. \] \[ R=18\,\text{mm}. \] \[ R=0.018\,\text{m}. \]
Step 2: Calculate the capacitance of the big drop. \[ C=4\pi\varepsilon_0R. \] Using \[ 4\pi\varepsilon_0=\frac1{9\times10^9}, \] \[ C=\frac{0.018}{9\times10^9}. \] \[ C=2\times10^{-12}\,\text{F}. \] \[ \boxed{C=2\,\text{pF}}. \]
Step 3: Find the total charge on the big drop. Each drop carries \[ q=10\,\text{nC}=10\times10^{-9}\,\text{C}. \] Hence, \[ Q=216q. \] \[ Q=216\times10\times10^{-9}. \] \[ Q=2.16\times10^{-6}\,\text{C}. \]
Step 4: Calculate the potential of the big drop. \[ V=\frac{Q}{C}. \] \[ V = \frac{2.16\times10^{-6}} {2\times10^{-12}}. \] \[ V = 1.08\times10^6\,\text{V}. \] Therefore, \[ \boxed{V=1.08\times10^6\,\text{V}} \] \[ \boxed{ \text{Capacitance}=2\,\text{pF}, \qquad \text{Potential}=1.08\times10^6\,\text{V} } \] \[ \boxed{\text{Answer = (C)}} \]
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