Two points charges +4qand q are kept at 6a distance apart. At what ppint on the line joining them the electric field intensity due to both the charge is 0
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∆I=I°-I.
∆I=0
I°=I.
kq•/x=kq./(6a-x)
4q/x=q/(6a-x)
x/4=6a-x
x=24a-4x
5x=24a
x=24a/5
so,the distance from +4q should be 24a/5
if +4q as origin the answer is ( 24a/5,0)
∆I=0
I°=I.
kq•/x=kq./(6a-x)
4q/x=q/(6a-x)
x/4=6a-x
x=24a-4x
5x=24a
x=24a/5
so,the distance from +4q should be 24a/5
if +4q as origin the answer is ( 24a/5,0)
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