Two unit negative charges are placed on a straight line. A positive charge q is placed exactly at the mid-point between these unit charges. If the system of three charges is in equilibrium the value of q (in C) is:

Question

Two unit negative charges are placed on a straight line. A positive charge q is placed exactly at the mid-point between these unit charges. If the system of three charges is in equilibrium the value of q (in C) is: (A) 1.0 (B) 0.75 (C) 0.5 (D) 0.25

Tardigrade Admin · Accepted Answer

For equilibrium of charge q. (1/4π ε 0)(-O× q/(r/2)2)+(1/4π ε 0)((-Q)(-Q)/r2)=0 ⇒ 4q=Q ⇒ a=(Q/4)=(1/4) (∵ Q=1 C) =0.25 C

Q. Two unit negative charges are placed on a straight line. A positive charge q is placed exactly at the mid-point between these unit charges. If the system of three charges is in equilibrium the value of q (in C) is:

1.0

0.75

0.5

0.25

For equilibrium of charge q.
$\frac{1}{4 π ε _{0}} \frac{- O \times q}{( r /2 ) ^{2}} + \frac{1}{4 π ε _{0}} \frac{( - Q ) ( - Q )}{r ^{2}} = 0$ $\Rightarrow$ $4 q = Q$ $\Rightarrow$ $a = \frac{Q}{4} = \frac{1}{4}$ $(∵ Q = 1 C)$ $= 0.25 C$

Q. Two unit negative charges are placed on a straight line. A positive charge q is placed exactly at the mid-point between these unit charges. If the system of three charges is in equilibrium the value of q (in C) is:

1.0

0.75

0.5

0.25

For equilibrium of charge q. 4πε0​1​(r/2)2−O×q​+4πε0​1​r2(−Q)(−Q)​=0 ⇒ 4q=Q ⇒ a=4Q​=41​ (∵Q=1C) =0.25C

For equilibrium of charge q.
$\frac{1}{4 π ε _{0}} \frac{- O \times q}{( r /2 ) ^{2}} + \frac{1}{4 π ε _{0}} \frac{( - Q ) ( - Q )}{r ^{2}} = 0$ $\Rightarrow$ $4 q = Q$ $\Rightarrow$ $a = \frac{Q}{4} = \frac{1}{4}$ $(∵ Q = 1 C)$ $= 0.25 C$