1. An oil drop has a mass of 4.95times 10^-15 kg experiencing a downward acceleration of 18.4m/s^2 Calculate the electric force.

Here's how to calculate the electric force on the oil drop:<br /><br />1. **Calculate the force due to gravity (weight):**<br /><br /> Weight (W) = mass (m) * acceleration due to gravity (g)<br /><br /> W = $(4.95 \times 10^{-15} kg) * (9.8 m/s^2)$<br /><br /> W = $4.851 \times 10^{-14} N$ (downward)<br /><br />2. **Calculate the net force:**<br /><br /> The oil drop is accelerating downwards, meaning the net force is downwards. The net force is the difference between the weight and the upward electric force (Fe).<br /><br /> Net Force (Fnet) = mass (m) * acceleration (a)<br /><br /> Fnet = $(4.95 \times 10^{-15} kg) * (18.4 m/s^2)$<br /><br /> Fnet = $9.108 \times 10^{-14} N$ (downward)<br /><br />3. **Calculate the electric force:**<br /><br /> Since the net force is downwards and greater than the weight, the electric force must be acting downwards as well. We can find the magnitude of the electric force by:<br /><br /> Fnet = W + Fe<br /><br /> Fe = Fnet - W<br /><br /> Fe = $(9.108 \times 10^{-14} N) - (4.851 \times 10^{-14} N)$<br /><br /> Fe = $4.257 \times 10^{-14} N$ (downward)<br /><br />Therefore, the electric force on the oil drop is $\mathbf{4.26 \times 10^{-14} N}$ **downwards**.<br />