Answer
(a) $1.3\times 10^6N/C $
(b) Less than
Work Step by Step
(a) The electric field at the mid point can be calculated as follows:
$ E=\frac{Kq}{h^2}$
We plug in the known values to obtain:
$ E=\frac{(8.99\times 10^9N.m/C^2)(4.7\times 10^{-6}C)}{(0.21m)^2-(\frac{0.21}{2})^2}$
$ E=1.3\times 10^6N/C $
(b) We know that the magnitude of the electric field is zero because of the symmetry and hence it is less than the field at the mid point of each side.
