the electric field due to point charge a distance 6 m from it is 630N/C the magnitude of charge is
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Answered by
13
E.F=KQ/R²
630=KQ/36
KQ=630×36
Q=(630×36)/K
630=KQ/36
KQ=630×36
Q=(630×36)/K
Answered by
26
Here's your answer--------------
As we know that,electric field are affected by inverse-square function,as distance from charge increases in a linear fashion,the field diminishes exponentially ,and equation is :
E=kq/d^2
Given,E=630N/C
k=9.0×10^9
d=6m
so,q=Ed^2/k
on putting the values
q=630×(6)^2/9.0×10^9
q=2.52×10^-6
【hope this will help ya】☺️
As we know that,electric field are affected by inverse-square function,as distance from charge increases in a linear fashion,the field diminishes exponentially ,and equation is :
E=kq/d^2
Given,E=630N/C
k=9.0×10^9
d=6m
so,q=Ed^2/k
on putting the values
q=630×(6)^2/9.0×10^9
q=2.52×10^-6
【hope this will help ya】☺️
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