2 point charges +16μ C and -9μ C are placed 8 cm apart in air . You are asked to place a +10μ C charge at a third position . Such that the net force on +10μ C charge is zero?
Answers
Explanation:
Given :
Two point charges 16μC and -9μC are placed at 8cm apart in air.
To find :
Determine the Position of 10μC charge such that force on this will be zero .
Theory :
•Coulombs law
\bf\:F =\dfrac{kq_{1}q_{2} }{r {}^{2} }F=
r
2
kq
1
q
2
Solution :
Let the point A has the positive charge +16 μC and the point B has the negative charge -9 μC.
Let the position of 10μC be P , x cm distance from point B .
P is the point where the electric force is zero.
The electric force acting on point P due to the positive charge at A is,
\bf\:F_{PA}=\sf\:\frac{k\times10\times16\times10{}^{-6}\times10{}^{-6}}{(x+8){}^{2}\times10{}^{-4}}F
PA
=
(x+8)
2
×10
−4
k×10×16×10
−6
×10
−6
= \frac{k \times 10 {}^{ - 12} \times 10 \times 16 }{(x + 8) {}^{2} \times 10 {}^{ - 4} }=
(x+8)
2
×10
−4
k×10
−12
×10×16
The electric force acting on point P due to the negative charge at B is,
\bf\:F_{PB}=\sf\:\frac{k\times10\times9\times10{}^{-6}\times10{}^{-6}}{(8){}^{2}\times10{}^{-4}}F
PB
=
(8)
2
×10
−4
k×10×9×10
−6
×10
−6
= \frac{k \times 10 {}^{ - 12} \times 9 \times 10 }{8 {}^{2} \times 10 {}^{ - 4} }=
8
2
×10
−4
k×10
−12
×9×10
Since force on 10μC at point P is zero
\implies\:\sf\:F_{PA}=F_{PB}⟹F
PA
=F
PB
\dfrac{k \times10 {}^{ - 12} \times10 \times 16 }{(x + 8) {}^{2} \times 10 { }^{ - 4} } = \dfrac{k \times 10 {}^{ - 12} \times 10 \times 9 }{8 {}^{2} \times 10 {}^{ - 4} }
(x+8)
2
×10
−4
k×10
−12
×10×16
=
8
2
×10
−4
k×10
−12
×10×9
\implies \dfrac{16}{(x + 8) {}^{2} } = \dfrac{9}{x {}^{2} }⟹
(x+8)
2
16
=
x
2
9
Taking square root on both sides
\implies \frac{4}{x + 8} = \frac{3}{x}⟹
x+8
4
=
x
3
\implies3(x + 8) = 4x⟹3(x+8)=4x
\implies \bf x = 24cm⟹x=24cm
_______________________________
Therefore,the Position of 10μC charge such that force on this will be zero is 24cm from -9μC charge .
