A particle travels in a circle of diameter 1.0 cm under the...

When the path of the particle is straight under the combined electric and magnetic fields, the electric force balances the magnetic force.

$qE = qvB \Rightarrow v = \dfrac{E}{B}$

Substituting the given values $E = 200 \text{ V m}^{-1}$ and $B = 0.40 \text{ T}$:

$v = \dfrac{200}{0.40} = 500 \text{ m s}^{-1}$

When the particle travels in a circle under the influence of the magnetic field alone, the radius of the circular path is given by:

$r = \dfrac{mv}{qB}$

Rearranging for the charge-to-mass ratio:

$\dfrac{q}{m} = \dfrac{v}{rB}$

The diameter of the circle is $1.0 \text{ cm}$, so the radius is $r = 0.5 \text{ cm} = 5 \times 10^{-3} \text{ m}$.

Substituting the values of $v$, $r$, and $B$:

$\dfrac{q}{m} = \dfrac{500}{5 \times 10^{-3} \times 0.40}$

$\dfrac{q}{m} = \dfrac{500}{2 \times 10^{-3}} = 2.5 \times 10^5 \text{ C kg}^{-1}$