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 600π V/m acting in the upward direction is ________.

Fill in the blank with the correct answer from the options given below. (Take g = 10 m/s²)

$2 \times 10^{-5} C$

The correct answer is Option 2: $2 \times 10^{-5}\text{ C}$

Since the ball remains just suspended in oil, the net force on it is zero. Therefore, the upward electric force and buoyant force together balance the weight of the ball.

$qE + \rho_{oil}Vg = \rho_{ball}Vg$

$qE = (\rho_{ball}-\rho_{oil})Vg$

Given:

$\rho_{ball} = 8000\ \text{kg/m}^3,\quad \rho_{oil} = 800\ \text{kg/m}^3$

Radius of the ball:

$r = 0.5\text{ cm} = 5 \times 10^{-3}\text{ m}$

Volume of the sphere:

$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(5 \times 10^{-3})^3 = \frac{500}{3}\pi \times 10^{-9}\text{ m}^3$

Substituting the values:

$q(600\pi) = (8000-800)\times \frac{500}{3}\pi \times 10^{-9}\times 10$

$600q = 7200 \times \frac{500}{3}\times 10^{-8} = 1.2 \times 10^{-2}$

$q = \frac{1.2 \times 10^{-2}}{600} = 2 \times 10^{-5}\text{ C}$