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5 votes
5 votes
Force between two bodies having charges 5.0 x 102 uC and 1.0 x 102 uC separated by

distance of 0.50 m in air is
1.0 x 103N
0 3.0 x 102 N
1800 N
O2200 N

Force between two bodies having charges 5.0 x 102 uC and 1.0 x 102 uC separated by-example-1
User Alexandre
by
2.5k points

1 Answer

13 votes
13 votes

Answer:

1800 N

Step-by-step explanation:

We know that ,


\longrightarrow\sf Force = k (q_1 q_2)/(r^2)\\


\longrightarrow\sf Force = ( 9 * 10^9 * 5 * 10^2 * 1 * 10^2 * 10 ^(-12))/( ( 0.5)^2)\\


\longrightarrow\sf Force = ( 9 * 10^9 * 5 * 10^2 * 1 * 10^2 * 10 ^(-12))/( 25 * 10^(-2))


\longrightarrow\sf Force = 1800 \ N \\\\

** Edits are welcomed **

User Dmastylo
by
2.7k points