the electric field at (30,30) cm due to a charge of -8nC at the origin in N/C is
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first find distance between origjn and (30,30)
use distance formula ,
r =√{(30-0)^2+(30-0)^2}
=√(900+900)=30√2 cm
now,
Electric field at (30,30) =kQ/r^2
= 9 x 10^9 x (-8) x 10^-9/(30√2/100)^2
=-72 x 10^2/18
= -400 N/ C
use distance formula ,
r =√{(30-0)^2+(30-0)^2}
=√(900+900)=30√2 cm
now,
Electric field at (30,30) =kQ/r^2
= 9 x 10^9 x (-8) x 10^-9/(30√2/100)^2
=-72 x 10^2/18
= -400 N/ C
kamal27:
sry it is wrong
how ?
what is your answer
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