Two charges +4 e and +e are at a distance x apart. At what distance a charge q must be placed from charge +e so that it is in equilibrium?

Question

Two charges +4 e and +e are at a distance x apart. At what distance a charge q must be placed from charge +e so that it is in equilibrium? (A) x/2 (B) 2x/3 (C) x/3 (D) x/6

Tardigrade Admin · Accepted Answer

For equilibrium of q, |F1|=|F2| <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/4a379f7035def7b54a3786eb3e4b3172-.png" /> ⇒ x2=(x/√(Q1)Q2+1)=(x/√(4 e)e+1)=(x/3)

Q. Two charges $+ 4 e$ and $+ e$ are at a distance $x$ apart. At what distance a charge $q$ must be placed from charge $+ e$ so that it is in equilibrium?

x/2

2x/3

x/3

x/6

For equilibrium of $q$ ,
$∣ F_{1} ∣ = ∣ F_{2} ∣$

$\Rightarrow x_{2} = \frac{x}{\frac{Q _{1}}{Q _{2}} + 1} = \frac{x}{\frac{4 e}{e} + 1} = \frac{x}{3}$

Q. Two charges +4e and +e are at a distance x apart. At what distance a charge q must be placed from charge +e so that it is in equilibrium?

x/2

2x/3

x/3

x/6

For equilibrium of q, ∣F1​∣=∣F2​∣ ⇒x2​=Q2​Q1​​​+1x​=e4e​​+1x​=3x​

Q. Two charges $+ 4 e$ and $+ e$ are at a distance $x$ apart. At what distance a charge $q$ must be placed from charge $+ e$ so that it is in equilibrium?

For equilibrium of $q$ ,
$∣ F_{1} ∣ = ∣ F_{2} ∣$

$\Rightarrow x_{2} = \frac{x}{\frac{Q _{1}}{Q _{2}} + 1} = \frac{x}{\frac{4 e}{e} + 1} = \frac{x}{3}$