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Q. Two metal plates having a potential difference of $800\, V$ and $2\, cm$ apart. It is found that a particle of mass $1.96 \times 10^{-15} \,kg$ remaining suspended in the region between the plates. The charge on the narticle must be $(e=$ elementry charge)

JIPMERJIPMER 2003Electrostatic Potential and Capacitance

Solution:

$E=\frac{V}{d}=\frac{800}{2 \times 10^{-2}} \,V / m $
$=4 \times 10^{4} \,V / m$
again, $q E=m g$
$q=\frac{m g}{E}=\frac{1.96 \times 10^{-15} \times 9.8}{4 \times 10^{4}} e$
$=3 e$