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Gravitational field charge question?
In this question, a charged drop of water experiences a gravitational field, and an electric field to stop it falling.
A small water drop with a charge of 110 times that of the electron is prevented from falling (in a vacuum) by an electric field E = 1.5×105V m−1 .
What is the direction of E ?
Select one:
down
up
What is the mass of the drop ?
Give your answer in kg. Remember that the acceleration due to gravity, g, is 9.8m s−2 , and the charge on a single electron is −1.6×10−19C .
1 Answer
- 5 years ago
The drop is negatively charged and the force is upwards, The direction of the field is the direction a positive test charge would tend to move. The electric field is therefore down.
Upward force on drop = E * q = (-1.5 * 10^5 * 110 * -1.6 * 10^-19) = 2.64 * 10^-12 newtons.
If the drop is stationary, the net force on the drop is zero.
So the magnitude of the gravitational (weight) force on the drop (mg) is also 2.64 * 10^-12 newtons
m = (2.64 * 10^-12 / 9.8) = 2.7 * 10^-13 kg
