Charges + q and - q are placed at points A and B respectively which are a distance 2 L apart, C is the mid point of A and B. The work done in moving a charge + Q along the semicircle CRD is :-

Question

Charges + q and - q are placed at points A and B respectively which are a distance 2 L apart, C is the mid point of A and B. The work done in moving a charge + Q along the semicircle CRD is :- (A) -( qQ /6 π ∈0 L ) (B) ( qQ /4 π ∈0 L ) (C) ( qQ /2 π ∈0 L ) (D) ( qQ /6 π ∈0 L )

Tardigrade Admin · Accepted Answer

Ui=(k q Q/L)-(k Q q/L)=0 Uf=(k q Q/3 L)-(k Q q/L)=-(2/3) (k Q q/L) So, W=Uf-Ui=-(2/3) (k Q q/L)=-(Q q/6 π ε0 L)

Q. Charges $+ q$ and $- q$ are placed at points $A$ and B respectively which are a distance $2 L$ apart, $C$ is the mid point of $A$ and $B$ . The work done in moving a charge $+ Q$ along the semicircle CRD is :-

$- \frac{qQ}{6 π \in _{0} L}$

$\frac{qQ}{4 π \in _{0} L}$

$\frac{qQ}{2 π \in _{0} L}$

$\frac{qQ}{6 π \in _{0} L}$

$U_{i} = \frac{k qQ}{L} - \frac{k Qq}{L} = 0$
$U_{f} = \frac{k qQ}{3 L} - \frac{k Qq}{L} = - \frac{2}{3} \frac{k Qq}{L}$
So, $W = U_{f} - U_{i} = - \frac{2}{3} \frac{k Qq}{L} = - \frac{Qq}{6 π ε _{0} L}$

Q. Charges +q and −q are placed at points A and B respectively which are a distance 2L apart, C is the mid point of A and B. The work done in moving a charge +Q along the semicircle CRD is :-

−6π∈0​LqQ​

4π∈0​LqQ​

2π∈0​LqQ​

6π∈0​LqQ​