The answer is c you got to look for answers that make sense
65 years but anything can happen to them
I’m not really sure but I hope this helps
Answer:
Somewhere between the two wires, but closer to the wire carrying λ₂
Explanation:
Electric Field for a point at distance x from an electric charge Q is Ef = K*Q/x².
Electric Fied due to an electric charge is a vector and its direction is such that if we place a positive charge in the point it will be rejected ( equal sign charge repulse each other and different attract each other)
According to that previous explanation, it is no possible two have Ef=0 out of the two wires region, since above the upper wire and below the lower wire we have to add the two electric fields (both have the same direction). Therefore we only have possibilities of Ef = 0 inside the two wires, where the repulsion produced over a positive charge due to the two wires are opposite
In the particular case in which λ₁ and λ₂ are equals then all the points exactly in the middle of d (distance between the two wires ) will have Ef =0.
As we can see at the beginning of the step by step explanation Electric field is proportional to the electric charge, or for a bigger charge, bigger Ef (keeping constant distance). In our case λ₁ >λ₂ then E₁ (Electric field produced by a wire carrying λ₁ will be bigger than (Electric field produced by wire carrying λ₂ at the middle way between the wires.
But for points closer to wire with λ₂ ( where E₂ is bigger than E₁ ) we will surely find an appropriate distance to get equals E and then Ef = 0
The potential at a distance r from a charge Q is given by

where ke is the Coulomb's constant.
The charge in our problem is

; for the point at

, the potential is

For the point at infinity, we immediately see that the potential is zero, because

and so

.
Therefore, the potential difference between the two points is