Explanation:
(a) E = F/q
E = 4.8×10^-17/1.6×10^-19
E = 300 N/C
(b) same magnitude of electric field is exerted on proton
a). The magnitude along with the direction of the electric field releasing westward force of [tex]4.80[/tex] × [tex]10^{-17}[/tex] N would be:
[tex]3[/tex] × [tex]10^{-36}[/tex] N/C is Eastward Direction
b). The magnitude along with the force of the direction that this field releases on proton would be:
[tex]4.8[/tex] × [tex]10^{17} N[/tex] in Eastward Direction
a). Given that,
Force [tex]=[/tex] [tex]4.80[/tex] × [tex]10^{-17}[/tex] N
As we know,
Force [tex]=[/tex] Charge × Electric Field
So,
∵ Electric Field [tex]= Force/Charge[/tex]
[tex]= 4.8[/tex] × [tex]10^{17}[/tex])[tex]/(1.6[/tex] × [tex]10^{-19}[/tex][tex])[/tex]
[tex]= 3[/tex] × [tex]10^{36}N/C[/tex]
The direction of the field would be opposite i.e. Eastward direction due to the field carrying a -ve charge.
b). The magnitude carried by the force working on the proton would be the same with an opposite direction due to +ve charge.
∵ Force [tex]=[/tex] [tex]4.80[/tex] × [tex]10^{-17}[/tex] N in Eastward direction.
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