1. Electric Charges and Fields
easy

A drop of ${10^{ - 6}}\,kg$ water carries ${10^{ - 6}}\,C$ charge. What electric field should be applied to balance its weight (assume $g = 10\,m/{s^2}$)

A

$10\, V/m$ upward

B

$10\, V/m$ downward

C

$0.1\, V/m$ downward

D

$0.1\, V/m$ upward

Solution

(a) By using $QE = mg$
$==>$ $E = \frac{{mg}}{Q} = \frac{{{{10}^{ – 6}} \times 10}}{{{{10}^{ – 6}}}} = 10\,V/m;$ upward because charge is positive.

Standard 12
Physics

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