An oil drop ‘B’ has charge 1.6 × 10−19 C and mass 1.6 × 10−14 kg. If the drop is in equilibrium position, let 10 k be the potential diff. between the plates. [The distance between the plates is 100 mm] Then what is the value of k.
To find the value of k, we need to consider the equilibrium position of the oil drop 'B' in an electric field between the plates.
In this scenario, the electric force acting on the oil drop is balanced by the gravitational force, maintaining equilibrium. The electric force (Fe) on the oil drop is given by the equation:
Fe = qE
Where:
q = charge of the oil drop = 1.6 × 10^(-19) C
E = electric field strength between the plates
The gravitational force (Fg) on the oil drop is given by the equation:
Fg = mg
Where:
m = mass of the oil drop = 1.6 × 10^(-14) kg
g = acceleration due to gravity = 9.8 m/s^2
In equilibrium, the electric force Fe and gravitational force Fg are equal, so we have:
qE = mg
Rearranging the equation, we can solve for the electric field E:
E = (mg) / q
Now, we know that the potential difference (V) between the plates is given as 10 kV (10,000 V) and the distance between the plates (d) is 100 mm (0.1 m). The electric field E between the plates can be related to potential difference by the equation:
E = V / d
Substituting the given values:
V / d = (mg) / q
Simplifying further:
10,000 / 0.1 = (1.6 × 10^(-14) × 9.8) / 1.6 × 10^(-19)
Dividing both sides by 1.6 × 10^(-19):
10,000 / 0.1 / (1.6 × 10^(-19)) = (1.6 × 10^(-14) × 9.8)
Simplifying the equation:
k = (1.6 × 10^(-14) × 9.8) / (10,000 / 0.1)
Calculating the above expression will give us the value of k.