Question

mercury from a metal vessel maintained at a potential of 3V falls in spherical drops of 2mm diameter through a small hole into a thin walled metal sphere of diameter 8cm , isolated and placed in air until the sphere is filled with mercury . calculate the final potential of the metal sphere

Answer 1

The final potential of the metal sphere is 4800 V.

Answer 2

The problem describes mercury drops falling from a vessel at a constant potential into an initially uncharged, isolated metal sphere until the sphere is filled. We need to find the final potential of the sphere.

1. Charge of a single drop: The mercury vessel is maintained at a potential $V_v = 3V$. When a drop is about to detach from the vessel, it is essentially part of the vessel and is at the same potential $V_v$. The drop is spherical with diameter $d = 2mm$, so its radius is $r = d/2 = 1mm = 10^{-3} m$. The capacitance of a small isolated conducting sphere of radius $r$ is $C_d = 4\pi\epsilon_0 r$. The charge on a single drop just before it detaches is $q = C_d V_v = (4\pi\epsilon_0 r) V_v$.

2. Number of drops filling the sphere: The sphere has diameter $D = 8cm$, so its radius is $R = D/2 = 4cm = 4 \times 10^{-2} m$. The volume of a single drop is $v_d = \frac{4}{3}\pi r^3$. The volume of the sphere is $V_s = \frac{4}{3}\pi R^3$. Let $N$ be the number of drops required to fill the sphere. The total volume of $N$ drops equals the volume of the sphere: $N \cdot v_d = V_s$ $N \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$ $N = \frac{R^3}{r^3} = \left(\frac{R}{r}\right)^3$ Substituting the values: $N = \left(\frac{4 \times 10^{-2} m}{10^{-3} m}\right)^3 = \left(\frac{40}{1}\right)^3 = 40^3 = 64000$.

3. Total charge accumulated on the sphere: Each drop carries a charge $q$. As the drops fall into the sphere and merge with the mercury inside, their charge is transferred to the sphere. The total charge $Q$ accumulated on the sphere when it is full is the sum of the charges of all $N$ drops: $Q = N \cdot q$ $Q = \left(\frac{R}{r}\right)^3 \cdot (4\pi\epsilon_0 r) V_v$ $Q = \frac{R^3}{r^3} \cdot 4\pi\epsilon_0 r V_v = 4\pi\epsilon_0 \frac{R^3}{r^2} V_v$.

4. Final potential of the sphere: When the sphere is full of mercury, it is a conducting sphere of radius $R$ carrying a total charge $Q$. The sphere is isolated. The capacitance of an isolated conducting sphere of radius $R$ is $C_s = 4\pi\epsilon_0 R$. The final potential $V_f$ of the sphere is given by $V_f = \frac{Q}{C_s}$. $V_f = \frac{4\pi\epsilon_0 \frac{R^3}{r^2} V_v}{4\pi\epsilon_0 R}$ $V_f = \frac{R^3}{r^2 R} V_v = \frac{R^2}{r^2} V_v = \left(\frac{R}{r}\right)^2 V_v$.

5. Substitute values and calculate: $R = 4 \times 10^{-2} m$ $r = 10^{-3} m$ $V_v = 3 V$ $V_f = \left(\frac{4 \times 10^{-2}}{10^{-3}}\right)^2 \times 3 V$ $V_f = \left(\frac{40}{1}\right)^2 \times 3 V$ $V_f = 40^2 \times 3 V$ $V_f = 1600 \times 3 V$ $V_f = 4800 V$.