Electric plates are 1.25cm apart, An o drop that weighs 1.1968x10^13 N is suspended between the plates when the potential differences across them is 550.0 volts.
What is the electric field intensity between the plates?
E = 550 V/0.0125m = 44,000 V/m
The force on the particle is not needed to compute the field, but can be used to compute the charge on the oil (?) drop. Why do you call it an o drop?
Sorry , didn't realized I just put o.
It's really suppose to be oil.
But then how do I go about finding how many excess electrons are on the oil drop?
To calculate the electric field intensity between the plates, we can use the formula:
Electric field intensity (E) = Voltage (V) / Distance (d)
In this case, the voltage between the plates is given as 550.0 volts, and the distance between the plates is 1.25 cm.
First, we need to convert the distance from centimeters to meters. Since 1 meter is equal to 100 centimeters, the distance between the plates is 1.25 cm * (1 meter / 100 centimeters) = 0.0125 meters.
Next, we can substitute the values into the formula:
E = 550.0 volts / 0.0125 meters
Calculating this gives us:
E = 44000 volts/meter
Therefore, the electric field intensity between the plates is 44000 volts/meter.