The charges $+ 4 q, - q$ and $+ 4 q$ are kept on a straight line at position $(0, 0, 0), (a, 0, 0)$ and $(2 a, 0, 0)$ respectively. Considering that they are free to move along the $x - a x i s$ only

All the charges are in stable equilibrium

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All the charges are in unstable equilibrium

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Only the middle charge is in stable equilibrium

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Only the middle charge is in unstable equilibrium

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Solution

The correct option is A All the charges are in stable equilibrium
Attraction between, $(+ 4 q a n d - q), F_{1} = \frac{k (+ 4 q) (- q)}{a^{2}} = 4 \frac{k q^{2}}{a^{2}}$

Repulsion between, $(+ 4 q a n d + 4 q), F_{2} = \frac{k (+ 4 q) (+ 4 q)}{{(2 a)}^{2}} = 4 \frac{k q^{2}}{a^{2}}$

Net repulsion and attraction force is zero,

Hence, System is in equilibrium

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Similar questions

Three charges +4q,-q and +4q are kept on a straight line at position (0,0,0),(a,0,0) and (2a,0,0) respectively. Considering that they are free to move along the x-axis only

Point charges +4q, -q and +4q are kept on the X-axis at points x=0,x=a and x=2a respectively. Are all the charges stable or unstable?

Q. Point charges +4q, -q, +4q are kept on the x-axis at points x=0,
x=a,x=2a respectively...

OPTIONS
1. Only -q is in stable equilibrium
2. All the charges are in stable eqbm
3.all the charges are in unstable eqbm
4. None of the charges are in eqbm

Sir ,please ..do provide the explanation or steps alongwith the right option ..so it's easily understood.

Q. Two point charges

+ q

and

+ q

are fixed at

(a, 0, 0)

and

(- a, 0, 0)

. A third point charge

- q

is at origin. State whether its equilibrium is stable, unstable or neutral if it is slightly displaced along

x -

axis.

Q. Two free point charges +q and +4q are at the co-ordinates (0,0) and (a, 0). A third charge is placed on the line joining them so that the entire system is in equilibrium. The value and position of the third charge is

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The charges + 4q,−q and +4q are kept on a straight line at position (0,0,0),(a,0,0) and (2a,0,0) respectively. Considering that they are free to move along the x−axis only

The correct option is A All the charges are in stable equilibriumAttraction between, (+4qand−q), F1=k(+4q)(−q)a2=4kq2a2 Repulsion between, (+4qand+4q),F2=k(+4q)(+4q)(2a)2=4kq2a2 Net repulsion and attraction force is zero, Hence, System is in equilibrium

The charges $+ 4 q, - q$ and $+ 4 q$ are kept on a straight line at position $(0, 0, 0), (a, 0, 0)$ and $(2 a, 0, 0)$ respectively. Considering that they are free to move along the $x - a x i s$ only