what is the magnitude of the potential difference δv14?

2.22×104V

Explanation:

The relationship between an uniform electric field →E, potential difference between points AandB (Δϕ), and distance d between them is:
→E=−Δϕd
.
−ve sign indicates that a positive charge will be repelled away from the positively charged plate, towards higher-voltage. This expression can be re-written as

Δϕ=−→E⋅→d
or Δϕ=−∣∣∣→E∣∣∣∣∣∣→d∣∣∣cosθ
Assuming that the electric field has component only along the distance vector and also it is uniform, cosθ=cos0=1. We obtain

|Δϕ|=∣∣∣→E∣∣∣∣∣∣→d∣∣∣
Inserting given values after converting distance in to meters,
Δϕ=3.7×107×0.60103
=2.22×104V