Question:
The magnitude of electric field intensity $E$, such that an electron placed in it would experience an electrical force equal to its weight, is given by
The magnitude of electric field intensity $E$, such that an electron placed in it would experience an electrical force equal to its weight, is given by
Updated On: Jun 18, 2022
- $m\,g\,e$
- $\frac{mg}{e}$
- $\frac{e}{mg}$
- $\frac{e^2}{m^2} g $
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The Correct Option is B
Solution and Explanation
Force on electron
$| F |= qE = eE = mg$
$E =\frac{ mg }{ e }$
$| F |= qE = eE = mg$
$E =\frac{ mg }{ e }$
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).