Answer
(a) $4.13\times 10^3N/C$
(b) $6.22\times 10^6m/s$
Work Step by Step
(a) The magnitude of the electric field can be determined as follows:
$E=\frac{m}{q}(\frac{2\Delta y(v^2)}{x^2})$
We plug in the known values to obtain:
$E=\frac{(9.11\times 10^{-31}Kg)2(0.618\times 10^{-2}m)(5.45\times 10^6m/s)^2}{(1.6\times 10^{-19}C)(2.25\times 10^{-2}m)^2}$
This simplifies to:
$E=4.13\times 10^3N/C$
(b) We can determine the required speed as follows:
$v_f=\sqrt{v_x^2+v_y^2}$
$\implies v_f=\sqrt{v^2+(\frac{2(\Delta y)v}{x})^2}$
$\implies v_f=v(\sqrt{1+4(\frac{\Delta y}{x})^2})$
We plug in the known values to obtain:
$v_f=(5.45\times 10^6m/s)\sqrt{1+4(\frac{0.00618m}{0.0225m})^2}=6.22\times 10^6m/s$
