Question:
An oil drop of charge q and mass m is in equilibrium in an electric field E. The charge of the oil drop is
An oil drop of charge q and mass m is in equilibrium in an electric field E. The charge of the oil drop is
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This is the fundamental principle behind Millikan's oil-drop experiment used to determine the elementary charge. Balancing forces is key to solving such equilibrium problems.
Updated On: Apr 21, 2026
- $q = \frac{mg}{E}$
- $q = \frac{E}{mg}$
- $q = \frac{mg}{2E}$
- $q = \frac{mg}{4E}$
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
For an oil drop to be in equilibrium (stationary or moving with constant velocity), the net force acting on it must be zero. The two primary vertical forces acting on the drop in this setup are the downward gravitational force and the upward electric force.
Step 2: Key Formula or Approach:
Gravitational force acting downwards: $F_g = mg$
Electric force acting upwards: $F_e = qE$
At equilibrium, these forces are equal in magnitude and opposite in direction.
\[ \Sigma F = F_e - F_g = 0 \]
Step 3: Detailed Explanation:
Equating the magnitudes of the two forces:
\[ F_e = F_g \]
\[ qE = mg \]
To find the charge $q$, rearrange the equation:
\[ q = \frac{mg}{E} \]
Step 4: Final Answer:
The correct option is (A) $q = \frac{mg}{E}$.
For an oil drop to be in equilibrium (stationary or moving with constant velocity), the net force acting on it must be zero. The two primary vertical forces acting on the drop in this setup are the downward gravitational force and the upward electric force.
Step 2: Key Formula or Approach:
Gravitational force acting downwards: $F_g = mg$
Electric force acting upwards: $F_e = qE$
At equilibrium, these forces are equal in magnitude and opposite in direction.
\[ \Sigma F = F_e - F_g = 0 \]
Step 3: Detailed Explanation:
Equating the magnitudes of the two forces:
\[ F_e = F_g \]
\[ qE = mg \]
To find the charge $q$, rearrange the equation:
\[ q = \frac{mg}{E} \]
Step 4: Final Answer:
The correct option is (A) $q = \frac{mg}{E}$.
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