1 × 10−6C
2 × 10−5C
To solve this problem, we want to find the charge on the copper ball such that it remains suspended in the oil. This means that the net force on the ball is zero. The forces acting on the copper ball are:
where:
The volume \(V\) of the sphere is given by:
\(V = \frac{4}{3}\pi r^3\) where \(r = 0.5 \, \text{cm} = 0.005 \, \text{m}\).
Thus,
\(V = \frac{4}{3}\pi (0.005)^3 = \frac{4}{3}\pi \times 1.25 \times 10^{-7} = \frac{5\pi}{3} \times 10^{-7} \, \text{m}^3\).
The condition for suspension is that the net upward force equals the downward force:
\((F_b + F_e) = F_g\).
Substituting the expressions for each force, we have:
\(V \cdot \rho_o \cdot g + qE = V \cdot \rho_c \cdot g\).
Rearranging for \(q\),
\(q = \frac{V(\rho_c - \rho_o)g}{E}\).
Substitute the given values:
Therefore,
\(q = \frac{\left(\frac{5\pi}{3} \times 10^{-7}\right)(8000 - 800) \cdot 10}{600\pi} = \frac{\frac{5\pi}{3} \times 10^{-7} \cdot 7200 \cdot 10}{600\pi}\).
This simplifies to:
\(q = \frac{5 \times 7200 \times 10^{-7} \cdot 10}{3 \times 600} = \frac{5 \times 7200 \times 10^{-6}}{1800}\).
\(= \frac{5 \times 4}{10} \times 10^{-6} = 2 \times 10^{-5} \, \text{C}\).
Thus, the correct answer is: \(2 \times 10^{-5} \text{C}\).