Answer
The strength of the electric field is $~~6.8\times 10^5~N/C$
Work Step by Step
To find the ion's speed, we can consider the kinetic energy after accelerating through the potential difference:
$K = q~V_1$
$\frac{1}{2}mv^2 = q~V_1$
$v^2 = \frac{2~q~V_1}{m}$
$v = \sqrt{\frac{2~q~V_1}{m}}$
$v = \sqrt{\frac{(2)(1.6\times 10^{-19})(10,000~V)}{9.99\times 10^{-27}~kg}}$
$v = 5.66\times 10^5~m/s$
The force on the ions due to the electric field must be equal in magnitude to the force on the ions due to the magnetic field.
We can find the magnitude of the electric field:
$qE = qvB$
$E = v~B$
$E = (5.66\times 10^5~m/s)(1.2~T)$
$E = 6.8\times 10^5~N/C$
The strength of the electric field is $~~6.8\times 10^5~N/C$
