A copper ball of density 8.0 g/cc and 1 cm in diameter is immersed in oil of density 0.8 g/cc. The charge on the ball if it remains just suspended in oil in an electric field of intensity \( 600\pi\,\mathrm{V/m} \) acting in the upward direction is _______________ .
Fill in the blank with the correct answer from the options given below. (Take \( g = 10\,\mathrm{m/s^2} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 2 \times 10^{-5}\,\mathrm{C} \)
The ball is in equilibrium: its weight acts downward, while the buoyant force of the oil and the electric force \( qE \) act upward. The field is upward, so the charge is positive.
\( \displaystyle qE + \sigma V g = \rho V g \;\Rightarrow\; q = \frac{(\rho - \sigma)\,V g}{E} \)
\( \rho = 8.0\,\mathrm{g/cc} = 8000\,\mathrm{kg\,m^{-3}} \)
\( \sigma = 0.8\,\mathrm{g/cc} = 800\,\mathrm{kg\,m^{-3}} \)
\( r = 0.5\,\mathrm{cm} = 5\times10^{-3}\,\mathrm{m} \)
\( \displaystyle V = \frac{4}{3}\pi r^{3} = \frac{4}{3}\pi\,(5\times10^{-3})^{3} = \frac{4}{3}\pi \times 1.25\times10^{-7}\,\mathrm{m^3} \)
\( \displaystyle q = \frac{7200 \times \frac{4}{3}\pi \times 1.25\times10^{-7} \times 10}{600\pi} \)
The \( \pi \) cancels:
\( \displaystyle q = \frac{7200 \times 1.667\times10^{-7} \times 10}{600} = \frac{1.2\times10^{-2}}{600} = 2\times10^{-5}\,\mathrm{C} \)