Three charges -q,- Q and +q are placed in a straight line as shown. If the total potential energy of the system is zero, then the ratio (q/Q) is

Question

Three charges -q,- Q and +q are placed in a straight line as shown. If the total potential energy of the system is zero, then the ratio (q/Q) is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/43fd8b36be72822f8b53afd233973308-.png" /> (A) 2 (B) 5.5 (C) 4 (D) 1.5

Tardigrade Admin · Accepted Answer

Potential energy of the system (-k q Q/x)-(k Q q/x)+(k q2/2 x)=0 ⇒ (-4 k q Q+k q2/2 x)=0 ⇒ k q2=4 k Q q ⇒ (q/Q)=4

Q. Three charges $- q, - Q$ and $+ q$ are placed in a straight line as shown. If the total potential energy of the system is zero, then the ratio $\frac{q}{Q}$ is

$2$

$5.5$

$4$

$1.5$

Potential energy of the system
$\frac{- k qQ}{x} - \frac{k Qq}{x} + \frac{k q ^{2}}{2 x} = 0$
$\Rightarrow \frac{- 4 k qQ + k q ^{2}}{2 x} = 0$
$\Rightarrow k q^{2} = 4 k Qq$
$\Rightarrow \frac{q}{Q} = 4$

Q. Three charges −q,−Q and +q are placed in a straight line as shown. If the total potential energy of the system is zero, then the ratio Qq​ is

2

5.5

4

1.5

Potential energy of the system x−kqQ​−xkQq​+2xkq2​=0 ⇒2x−4kqQ+kq2​=0 ⇒kq2=4kQq ⇒Qq​=4

Q. Three charges $- q, - Q$ and $+ q$ are placed in a straight line as shown. If the total potential energy of the system is zero, then the ratio $\frac{q}{Q}$ is

$2$

$5.5$

$4$

$1.5$

Potential energy of the system
$\frac{- k qQ}{x} - \frac{k Qq}{x} + \frac{k q ^{2}}{2 x} = 0$
$\Rightarrow \frac{- 4 k qQ + k q ^{2}}{2 x} = 0$
$\Rightarrow k q^{2} = 4 k Qq$
$\Rightarrow \frac{q}{Q} = 4$