91Ó°ÊÓ

$\stackrel{→}{\mathrm{B}}=-2.5\hat{\mathrm{i}}\mathrm{mT}$

$\stackrel{→}{\mathrm{v}}=2000\hat{\mathrm{j}}\mathrm{m}/\mathrm{s}$

The total force acting on the charged particle is the sum of the forces due to electric and magnetic fields.

Formulae are as follow:

Force acting on the charged particle due to electric field,

$F_e=qE$

Force acting on the charged particle due to magnetic field,

$F_m=qvB$

Where, $F_m$ is magnetic force, v is velocity, B is magnetic field, q is charge of particle.

The net force on the proton when $\stackrel{→}{E}=4.00\hat{\mathrm{k}}\mathrm{V}/\mathrm{m}$:

$$\begin{gathered} \stackrel{→}{\mathrm{F}}=\mathrm{q}\left(\stackrel{→}{\mathrm{E}}+\stackrel{→}{\mathrm{V}}×\stackrel{→}{\mathrm{B}}\right) \\ \stackrel{→}{\mathrm{F}}=1.6×{10}^{-19}\left(4.00\hat{\mathrm{k}}+2000\hat{\mathrm{j}}×\left(-2.5×{10}^{-3}\right)\hat{\mathrm{i}}\right) \\ \stackrel{→}{\mathrm{F}}=1.6×{10}^{-19}\left(4.00\hat{\mathrm{k}}+5.00\hat{\mathrm{k}}\right) \\ \stackrel{→}{\mathrm{F}}=1.44×{10}^{-18}\hat{\mathrm{k}}\mathrm{N} \end{gathered}$$

Hence, the net force acting on the proton is $\stackrel{→}{F}=1.44×{10}^{-18}\hat{k}\mathrm{N}$

Then net force on the proton when $\stackrel{→}{E}=-4.00\hat{\mathrm{k}}\mathrm{V}/\mathrm{m}$:

$$\begin{gathered} \stackrel{→}{\mathrm{F}}=\mathrm{q}\left(\stackrel{→}{\mathrm{E}}+\stackrel{→}{\mathrm{V}}×\stackrel{→}{\mathrm{B}}\right) \\ \stackrel{→}{\mathrm{F}}=1.6×{10}^{-19}\left(-4.00\hat{\mathrm{k}}+2000\hat{\mathrm{j}}×\left(-2.5×{10}^{-3}\right)\hat{\mathrm{i}}\right) \\ \stackrel{→}{\mathrm{F}}=1.6×{10}^{-19}\left(-4.00\hat{\mathrm{k}}+5.00\hat{\mathrm{k}}\right) \\ \stackrel{→}{\mathrm{F}}=1.60×{10}^{-19}\hat{\mathrm{k}}\mathrm{N} \end{gathered}$$

Hence, the net force acting on the proton is $\stackrel{→}{\mathrm{F}}=1.60×{10}^{-19}\hat{\mathrm{k}}\mathrm{N}$

The net force on the proton when $\stackrel{→}{E}=4.00\hat{\mathrm{i}}\mathrm{V}/\mathrm{m}$

$$\begin{gathered} \stackrel{→}{\mathrm{F}}=\mathrm{q}\left(\stackrel{→}{\mathrm{E}}+\stackrel{→}{\mathrm{V}}×\stackrel{→}{\mathrm{B}}\right) \\ \stackrel{→}{\mathrm{F}}=1.6×{10}^{-19}\left(4.00\hat{\mathrm{i}}+2000\hat{\mathrm{j}}×\left(-2.5×{10}^{-3}\right)\hat{\mathrm{i}}\right) \\ \stackrel{→}{\mathrm{F}}=1.6×{10}^{-19}\left(4.00\hat{\mathrm{i}}+5.00\hat{\mathrm{k}}\right) \\ \stackrel{→}{\mathrm{F}}=\left(6.41×{10}^{-19}\right)\hat{\mathrm{i}}\mathrm{N}+\left(8.01×{10}^{-19}\right)\hat{\mathrm{k}}\mathrm{N} \end{gathered}$$

Hence, the net force acting on the proton is $\stackrel{→}{\mathrm{F}}=\left(6.41×{10}^{-19}\right)\hat{\mathrm{i}}\mathrm{N}+\left(8.01×{10}^{-19}\right)\hat{\mathrm{k}}\mathrm{N}$.

Therefore, the values of net force due to different electric fields can be determined by taking the vector sum of forces due to the electric and magnetic field.