A) \[\frac{em}{g}\]
B) \[\frac{mg}{e}\]
C) emg
D) mg
Correct Answer: B
Solution :
Key Idea: A charge placed in an electric field experiences an electrostatic force. The intensity of electric field at a point in an electric field is the ratio of the force acting on the test charge placed at the point to the magnitude of test charge. \[\vec{E}=\frac{{\vec{F}}}{{{q}_{0}}}\] Also, the force acting on a charge 'q' placed at that point is \[F={{q}_{0}}E\] Also, F = mg = weight, where g is gravity. \[\therefore \] \[mg=eE\](given\[{{q}_{0}}=e\]) \[\therefore \] \[E=\frac{mg}{e}\]
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