# Question #8e4e9

Apr 15, 2017

$1.02 \times {10}^{8} \text{ V}$, rounded to two decimal places

#### Explanation:

Let the voltage required between the horizontal plates be $= V$
Electric field between the plates$= \frac{V}{d} = \frac{V}{0.1} = 10 V {\text{ Vm}}^{-} 1$

Force due to electric field on the proton$= q \vec{E}$
$= 1.6 \times {10}^{-} 19 \times 10 V$
$= 1.6 \times {10}^{-} 18 V \text{ N}$

Downwards force due to weight of proton$= m g$
$1.67 \times {10}^{-} 27 \times 9.81 = 1.638 \times {10}^{-} 26 \text{ N}$

To hold the proton motionless both forces must be balanced. We get
$1.6 \times {10}^{-} 18 V = 1.638 \times {10}^{-} 26$
$\implies V = \frac{1.638 \times {10}^{-} 26}{1.6 \times {10}^{-} 18}$
$\implies V = 1.02 \times {10}^{8} \text{ V}$, rounded to two decimal places