Question

List-I shows different configurations of two infinitely long charged wires, and List-II provides possible electric field magnitudes at a specific point. Match each entry in List-I with an appropriate entry from List-II, and choose the correct option. (k = 1/ $4\pi\epsilon_0$)

• A. Electric field at the midpoint is 6k$\lambda$/ d
• B. Electric field at the midpoint is 2 k$\lambda$/ d
• C. Electric field at the midpoint is 8 k$\lambda$/ d
• D. Electric field at the midpoint is 0

Answer 1

Correct matching: (P) – (4), (Q) – (3), (R) – (2), (S) – (1). Thus, none of the provided options is correct.

Answer 2

We begin by recalling that the electric field due to an infinitely long straight wire with charge per unit length $\lambda$ at a perpendicular distance r is

$E=\frac{2k\lambda}{r}$,

with k = 1/(4πε₀). At the midpoint of two wires separated by distance d the distance from each wire is r = d/2 so that the magnitude of the field from a wire having charge density λ is

$E=\frac{2k\lambda}{d/2}=\frac{4k\lambda}{d}$.

Now we analyze each configuration:

(P) Two wires, both with charge density +λ, separated by d

Place the wires at x = –d/2 and x = d/2. At the midpoint (0,0):

• Wire at x = –d/2 gives a field of magnitude 4kλ/d directed away from the wire → to the right.
• Wire at x = d/2 gives a field of the same magnitude 4kλ/d directed away from the wire → to the left.

Since they are equal and opposite, they cancel. Thus, field = 0 → matches List-II entry (4).

(Q) Two wires with charges +λ and –λ separated by d

Assume +λ is at x = –d/2 and –λ at x = d/2.

• Wire at x = –d/2 (positive) produces a field of 4kλ/d at the midpoint directed away from it → to the right.
• Wire at x = d/2 (negative) produces a field of 4kλ/d, but for a negative charge the field points toward the wire, i.e. also to the right.

They add up:

$E_{net} = \frac{4k\lambda}{d} + \frac{4k\lambda}{d} = \frac{8k\lambda}{d}$.

So (Q) matches List-II entry (3).

(R) Two wires with charges +λ and +λ/2 separated by d

Assume +λ is at x = –d/2 and +λ/2 is at x = d/2.

• Wire at x = –d/2 gives a field of 4kλ/d at the midpoint to the right.
• Wire at x = d/2 gives a field of

$E=\frac{2k(\lambda/2)}{d/2}=\frac{2k\lambda}{d}$,

directed away from it (i.e. to the left).

Subtracting (because the fields oppose):

$E_{net} = \frac{4k\lambda}{d} - \frac{2k\lambda}{d} = \frac{2k\lambda}{d}$.

Thus (R) matches List-II entry (2).

(S) Two wires with charges +λ and –λ/2 separated by d

Assume +λ at x = –d/2 and –λ/2 at x = d/2.

• Wire at x = –d/2: field of 4kλ/d at the midpoint to the right.
• Wire at x = d/2: field is

$\frac{2k(|-λ/2|)}{d/2}=\frac{2k(λ/2)}{d/2}=\frac{2k\lambda}{d}$,

and for a negative charge the field is toward the wire, i.e. from the midpoint to x = d/2, also to the right.

They add:

$E_{net} = \frac{4k\lambda}{d} + \frac{2k\lambda}{d} = \frac{6k\lambda}{d}$.

Thus (S) matches List-II entry (1).

Summary of correct matchings:

• (P) → (4)
• (Q) → (3)
• (R) → (2)
• (S) → (1)

Checking the provided answer choices:

• Option 1: (P, 3), (Q, 4), (R, 2), (S, 1)
• Option 2: (P, 4), (Q, 2), (R, 3), (S, 1)

Neither option matches our result (P,4), (Q,3), (R,2), (S,1).

Thus, none of the provided options is correct.