Question

What are the electric field strength and direction at the position of the electron in FIGURE$\mathrm{P}31.28$$?$

Short answer

Electric field strength in figure$\mathrm{P}31.8$is $1.75\times {10}^6\mathrm{V}/\mathrm{m}$,${\mathrm{F}}_{{\mathrm{E}}_{}}$isrole="math" localid="1648932697397" $2.8\times {10}^{-13}\mathrm{N}$and${\mathrm{F}}_{\mathrm{B}}$is$1.6\times {10}^{13}\mathrm{N}$

Step-by-step solution

Figure for direction of electric field intensity is depicted.

Figure is,

The figure on the page has come to an end is,

Calculation for force

Magnetic and electric fields both produce force:

$\tan \left(30°\right)=\frac{{\mathrm{F}}_{\mathrm{B}}}{{\mathrm{F}}_{\mathrm{E}}}$

A magnetic force occurs upward when utilizing the right hand rule.

${\mathrm{F}}_{\mathrm{B}}=\mathrm{qvB}$

${\mathrm{F}}_{\mathrm{B}}=1.6\times {10}^{-19}\times {10}^7\times 0.1$

${\mathrm{F}}_{\mathrm{B}}=1.6\times {10}^{-13}\mathrm{N}$

For ${\mathrm{F}}_{\mathrm{E}}$:

$\tan \left(30°\right)=\frac{{\mathrm{F}}_{\mathrm{B}}}{{\mathrm{F}}_{\mathrm{E}}}=\frac{1.6\times {10}^{13}}{{\mathrm{F}}_{\mathrm{E}}}$

${\mathrm{F}}_{\mathrm{E}}=\frac{1.6\times {10}^{-13}}{0.57}$

${\mathrm{F}}_{\mathrm{E}}=2.8\times {10}^{-13}\mathrm{N}$

Calculation for electric field strength

Strength of the electric field,

${\mathrm{F}}_{\mathrm{E}}=\mathrm{qE}$

$\mathrm{E}=\frac{{\mathrm{F}}_{\mathrm{E}}}{\mathrm{q}}$

${\mathrm{F}}_{\mathrm{E}}$-Force of electric field

$\mathrm{q}$-charge

$\mathrm{E}=\frac{2.8\times {10}^{-13}}{1.6\times {10}^{-19}}$

localid="1648932420433" $\mathrm{E}=1.75\times {10}^6\mathrm{V}/\mathrm{m}$