a) -10,800 N/C
b) [tex]4.79\cdot 10^{10}m/s^2[/tex]
c) [tex]4.88\cdot 10^9[/tex] times g
Explanation:
a)
The magnitude of the electric field produced by a single-point charge is given by
[tex]E=\frac{kQ}{r^2}[/tex]
where
k is the Coulomb's constant
Q is the charge producing the field
r is the distance at which the field is calculated
In this problem:
[tex]Q=-3.0\cdot 10^{-5}C[/tex] is the c harge producing the field
[tex]r=5.0 m[/tex] is the distance at which we want to calculate the field
Substituting,
[tex]E=\frac{(9.0\cdot 10^9)(-3.0\cdot 10^{-5})}{(5.0)^2}=-10,800 N/C[/tex]
where the negative sign indicates that the direction of the field is towards the charge producing the field (for a negative charge, the electric field is inward, towards the charge)
b)
The force experienced by a charged particle in an electric field is given by
[tex]F=qE[/tex]
where
q is the magnitude of the charge
E is the electric field
Moreover, the force on an object can be written as:
[tex]F=ma[/tex]
where
m is the mass
a is the acceleration
Combining the two equations,
[tex]ma=qE\\a=\frac{qE}{m}[/tex]
In this problem:
[tex]q=1.6\cdot 10^{-19}C[/tex] is the charge of the proton
[tex]E=0.50 kN/C=500 N/C[/tex] is the strength of the electric field
[tex]m=1.67\cdot 10^{-27} kg[/tex] is the mass of the proton
Substituting, we find the acceleration of the proton:
[tex]a=\frac{(1.6\cdot 10^{-19})(500)}{(1.67\cdot 10^{-27})}=4.79\cdot 10^{10}m/s^2[/tex]
c)
The acceleration due to gravity is the acceleration at which every object near the Earth's surface falls down, in absence of air resistance, and it is given by
[tex]g=9.81 m/s^2[/tex]
On the other hand, the acceleration of the proton in this problem is:
[tex]a=4.79\cdot 10^{10} m/s^2[/tex]
To find how many times greater is this acceleration than that due to gravity, we can divide the acceleration of the proton by the value of g. Doing so, we find:
[tex]\frac{a}{g}=\frac{4.79\cdot 10^{10}}{9.81}=4.88\cdot 10^9[/tex]
So, it is [tex]4.88\cdot 10^9[/tex] times greater than g.
