A charge Q is placed at the centre of the line joining two point charges +q and +q as shown in figure. The ratio of charges Q and q is

Question

A charge Q is placed at the centre of the line joining two point charges +q and +q as shown in figure. The ratio of charges Q and q is (A) 4 (B) 1/4 (C) -4 (D) -1/4

Tardigrade Admin · Accepted Answer

For the system to be in equilibrium, net force on q= 0 or (Qq/4πε0x2)+(qq/4πε0(2x)2)=0 or Q=(-q/4) or (Q/q)=-(1/4)

Q. A charge $Q$ is placed at the centre of the line joining two point charges $+ q$ and $+ q$ as shown in figure. The ratio of charges $Q$ and $q$ is

$4$

$1/4$

$- 4$

$- 1/4$

For the system to be in equilibrium,
net force on $q = 0$ or $\frac{Qq}{4 π ε _{0} x ^{2}} + \frac{qq}{4 π ε _{0} ( 2 x ) ^{2}} = 0$
or $Q = \frac{- q}{4}$ or $\frac{Q}{q} = - \frac{1}{4}$

Q. A charge Q is placed at the centre of the line joining two point charges +q and +q as shown in figure. The ratio of charges Q and q is

4

1/4

−4

−1/4

For the system to be in equilibrium, net force on q=0 or 4πε0​x2Qq​+4πε0​(2x)2qq​=0 or Q=4−q​ or qQ​=−41​

Q. A charge $Q$ is placed at the centre of the line joining two point charges $+ q$ and $+ q$ as shown in figure. The ratio of charges $Q$ and $q$ is

$4$

$1/4$

$- 4$

$- 1/4$

For the system to be in equilibrium,
net force on $q = 0$ or $\frac{Qq}{4 π ε _{0} x ^{2}} + \frac{qq}{4 π ε _{0} ( 2 x ) ^{2}} = 0$
or $Q = \frac{- q}{4}$ or $\frac{Q}{q} = - \frac{1}{4}$