Two isolated conducting spheres A and B have radii 3 cm and 6 cm and are far apart. Sphere A carries a charge of 9 nC and sphere B is uncharged. They are now joined by a thin conducting wire. The common potential of the two spheres is (take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(900\ \mathrm{V}\)
The wire joins the two spheres into one conductor. In the static state, a conductor has the same potential everywhere. So charge flows until both spheres are at one common potential. A sphere of radius \( R \) with charge \( q \) has potential \( q/4\pi\varepsilon_0R \).
Equal potentials give \( q_1/R_1=q_2/R_2 \). The total charge of 9 nC therefore divides in the ratio 3 : 6, so \( q_1=3\ \mathrm{nC} \) on the smaller sphere and \( q_2=6\ \mathrm{nC} \) on the larger one.
\( \begin{aligned} V &= \frac{q_1}{4\pi\varepsilon_0R_1} \\ &= \frac{9\times10^{9}\times3\times10^{-9}}{0.03} \\ &= 900\ \mathrm{V} \end{aligned} \)
The larger sphere gives the same value, \( 9\times10^{9}\times6\times10^{-9}/0.06 \). Therefore, the common potential is 900 V.
NCERT (2026-27):
Electrostatics of Conductors – “4. Electrostatic potential is constant throughout the volume of the conductor and has the same value (as inside) on its surface”
Potential Due to a System of Charges – “… potential is constant inside the shell (as no work is done in moving a charge inside the shell), and, therefore, equals its value at the surface, which is”
\( V=\frac{1}{4\pi\varepsilon_0}\frac{q}{R} \)\( \qquad [2.19(b)] \)