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Two fixed charges, +1.0 x 10^-6 C and -3.0 x 10^-6 C, are 10 cm apart. (a) Where may a third charge be located so that the net force acting on this charge is zero? (b) Is the equilibrium stable or not?

Respuesta :

Answer:

Part a)

the third charge will be placed at 13.66 cm on the other side of [tex]1.0 \times 10^{-6} C[/tex] charge

Part b)

If the charge is displaced by small distance towards left or right the force on it will move it away from its initial position

so this is Not stable equilibrium

Explanation:

Two charges are placed here on straight line are at 10 cm apart

here the two charges given are of opposite sign and hence the force on the third charge placed here on the same line will be zero where electric field is zero

Here electric field will be zero at the position near to the charge which is of small magnitude

so we will have

[tex]\frac{kq_1}{r^2} = \frac{kq_2}{(10 + r)^2}[/tex]

now we have

[tex]\frac{(1 \times 10^{-6})}{r^2} = \frac{(3\times 10^{-6})}{(10+ r)^2}[/tex]

[tex]10 + r = \sqrt3 r[/tex]

[tex]r = \frac{10}{\sqrt3 - 1} = 13.66 cm[/tex]

so the third charge will be placed at 13.66 cm on the other side of [tex]1.0 \times 10^{-6} C[/tex] charge

Part b)

If the charge is displaced by small distance towards left or right the force on it will move it away from its initial position

so this is Not stable equilibrium

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