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A change of $1 \mu C$ is divided into two parts such that their charges are in the ratio of $2:3$. These two charges are kept at a distance 1 m apart in vacuum. Then, the electric force between them (in Newtons) is

\[\begin {array} {1 1} (1)\;0.216 & \quad (2)\;0.00216 \\ (3)\;0.0216 & \quad (4)\;2.16 \end {array}\]

1 Answer

(2) 0.02216
answered Nov 7, 2013 by pady_1
 
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