Charges +q and -q are placed at A and B , respectively, which are at a distance 2L apart. If C is the midpoint between A and B , then work done in moving charge Q from C to D (through semicircular path CRD ) and from C to ∈ fty are, respectively,

Question

Charges +q and -q are placed at A and B , respectively, which are at a distance 2L apart. If C is the midpoint between A and B , then work done in moving charge Q from C to D (through semicircular path CRD ) and from C to ∈ fty are, respectively, <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-ryumi3kolcldiy6y.jpg" /> (A) (q Q/2 π ϵ 0 L) and infinity (B) (q Q/6 π ϵ 0 L) and zero (C) zero, zero (D) (- q Q/6 π ϵ 0 L) and zero

Tardigrade Admin · Accepted Answer

WC R D=Q(VD - VC) WC R D=Q[VD - 0] ( ∵ potential at centre of dipole is zero) WC R D=[(k q/3 L) - (k q/L)]Q WC R D=-(k q Q/L)× (2/3)=-(q Q/6 π ϵ 0 L)

Q. Charges $+ q$ and $- q$ are placed at $A$ and $B$ , respectively, which are at a distance $2 L$ apart. If $C$ is the midpoint between $A$ and $B$ , then work done in moving charge $Q$ from $C$ to $D$ (through semicircular path $CR D$ ) and from $C$ to $\in f t y$ are, respectively,

$\frac{qQ}{2 π ϵ _{0} L}$ and infinity

$\frac{qQ}{6 π ϵ _{0} L}$ and zero

zero, zero

$\frac{- qQ}{6 π ϵ _{0} L}$ and zero

$W_{CR D} = Q (V_{D} - V_{C})$
$W_{CR D} = Q [V_{D} - 0]$ ( $∵$ potential at centre of dipole is zero)
$W_{CR D} = [\frac{k q}{3 L} - \frac{k q}{L}] Q$
$W_{CR D} = - \frac{k qQ}{L} \times \frac{2}{3} = - \frac{qQ}{6 π ϵ _{0} L}$

Q. Charges +q and −q are placed at A and B , respectively, which are at a distance 2L apart. If C is the midpoint between A and B , then work done in moving charge Q from C to D (through semicircular path CRD ) and from C to ∈fty are, respectively,

2πϵ0​LqQ​ and infinity

6πϵ0​LqQ​ and zero