Point charges q , -q , 2Q and Q are placed in order at the corners A , B , C and D of a square of side 2b . If the field at the midpoint of CD is zero, then (q/Q) is

Question

Point charges q , -q , 2Q and Q are placed in order at the corners A , B , C and D of a square of side 2b . If the field at the midpoint of CD is zero, then (q/Q) is (A) 1 (B) 2 (C) (2 √2/5) (D) (5 √5/2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-z2tmpd15fmfuqxen.jpg" /> E=EC-ED-EAcos θ -EBcosâ¡θ (k (2 Q)/b2)-(k Q/b2)=2× (k q b/(b2 + 4 b2)(3)2) ⇒ Q=(2 q/5 √5)

Q. Point charges $q$ , $- q$ , $2 Q$ and $Q$ are placed in order at the corners $A$ , $B$ , $C$ and $D$ of a square of side $2 b$ . If the field at the midpoint of $C D$ is zero, then $\frac{q}{Q}$ is

$1$

$2$

$\frac{2 2}{5}$

$\frac{5 5}{2}$

$E = E_{C} - E_{D} - E_{A} cos θ - E_{B} cos θ$
$\frac{k ( 2 Q )}{b ^{2}} - \frac{k Q}{b ^{2}} = 2 \times \frac{k q b}{( b ^{2} + 4 b ^{2} ) ^{\frac{3}{2}}}$
$\Rightarrow Q = \frac{2 q}{5 5}$

Q. Point charges q , −q , 2Q and Q are placed in order at the corners A , B , C and D of a square of side 2b . If the field at the midpoint of CD is zero, then Qq​ is

1

2

522​​

255​​

E=EC​−ED​−EA​cosθ−EB​cos⁡θ b2k(2Q)​−b2kQ​=2×(b2+4b2)23​kqb​ ⇒Q=55​2q​

Q. Point charges $q$ , $- q$ , $2 Q$ and $Q$ are placed in order at the corners $A$ , $B$ , $C$ and $D$ of a square of side $2 b$ . If the field at the midpoint of $C D$ is zero, then $\frac{q}{Q}$ is

$1$

$2$

$\frac{2 2}{5}$

$\frac{5 5}{2}$

$E = E_{C} - E_{D} - E_{A} cos θ - E_{B} cos θ$
$\frac{k ( 2 Q )}{b ^{2}} - \frac{k Q}{b ^{2}} = 2 \times \frac{k q b}{( b ^{2} + 4 b ^{2} ) ^{\frac{3}{2}}}$
$\Rightarrow Q = \frac{2 q}{5 5}$