Question

What electric field strength and direction will allow the proton in FIGURE P31.30 to pass through this region of space without being deflected?

Short answer

The region of space without being deflected$E=2\times {10}^5\mathrm{V}/\mathrm{m}$

Step-by-step solution

Given

$v=2\times {10}^7\mathrm{m}/\mathrm{s}$

$B=0.01\mathrm{T}$

Right Hand Rule

Magnetic force occurs downward if using its right hand rule.

As a reason, due to the electric field, equal and opposite charges exist upward, but electron are not deflected. As a function, the direction of the electric field is also downward.

$F_e=F_B$

$qE=qvB$

$E=vB$

$E=2\times {10}^7\times {10}^{-2}$

$E=2\times {10}^5\mathrm{V}/\mathrm{m}$