2 point charges of +16 microcoulomb and -9 microcoulomb are placed 8 cm apart in air. Determine the position of the point where resultant electric field is 0 ( zero).
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Answered by
45
when electric field is zero
q/4π€r²= q/4π€r'²
9×10power9 ×16×10power-6/r² =9×10power9×9×10 power-6/(0.08-r)²
144/r²=81/(0.08-r)²
12/r=9/(0.08-r)
9r=0.96-12r
21r=0.98
r=4.6cm from 16 micro coloumb
q/4π€r²= q/4π€r'²
9×10power9 ×16×10power-6/r² =9×10power9×9×10 power-6/(0.08-r)²
144/r²=81/(0.08-r)²
12/r=9/(0.08-r)
9r=0.96-12r
21r=0.98
r=4.6cm from 16 micro coloumb
Answered by
19
Answer:
Explanation:
given: Distance between two charges is 8cm
let a point q is placed from distance x from -9mc
now, distance from +16mc to point charge q = [8+x] cm
force of electric field of both the charges must be equal
so that the resultant electric field is 0
F=k q / r^2
F1= k×16/ [8+x]^2 F2= k×9/ x^2
F1=F2
k×16/ [8+x]^2= k×9/x^2
square root on both sides
4/[8+x] = 3/ x
3 [8+x] = 4x
24+3x = 4x ⇒ 4x-3x = 24
x = 24
therefore, the position of point q where resultant electric field is zero is 24
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