A proton moves at 3.60 *10^5 m/s in the horizontal direction. It enters a uniform vertical electric field with a magnitude of 8.20* 10^3 N/C.
Find the time interval required for the proton to travel 4.00 cm horizontally.
thanks
To find the time interval required for the proton to travel 4.00 cm horizontally, we can use the equation of motion:
\[ s = ut + \frac{1}{2} a t^2 \]
where:
- \( s \) is the distance traveled horizontally (4.00 cm or 0.04 m),
- \( u \) is the initial velocity of the proton in the horizontal direction (3.60 * 10^5 m/s),
- \( t \) is the time interval we need to find, and
- \( a \) is the acceleration of the proton in the horizontal direction (which is zero since there is no horizontal force acting on the proton).
In this case, since there is no horizontal force acting on the proton, the proton will continue to move along the horizontal direction at a constant speed, so the acceleration \( a \) is zero.
Using the equation with \( a = 0 \), we can simplify it to:
\[ s = ut \]
Rearranging the equation to solve for \( t \), we have:
\[ t = \frac{s}{u} \]
Now let's substitute the values into the equation:
\[ t = \frac{0.04 \, \text{m}}{3.60 \times 10^5 \, \text{m/s}} \]
Calculating this, we get:
\[ t = 1.11 \times 10^{-7} \, \text{s} \]
Therefore, the time interval required for the proton to travel 4.00 cm horizontally is approximately \( 1.11 \times 10^{-7} \) seconds.