Answer:
The work done by the electric force is = [tex]+2.20*10^{-6} J[/tex]
The potential of the starting point with respect to the end point is [tex]= 523.8 V[/tex]
The magnitude of E [tex]= 8730 \ N/C[/tex]
Explanation:
Looking at this we can see that the work done by the electric force is equal to the kinetic energy this is because on the electric field is uniform and that the work done resulted in motion.
Mathematically
[tex]\Delta V q = W[/tex]
Where [tex]\Delta V[/tex] is the potential difference
q is the charge whose value i given as = [tex]4.20 *10 ^{-9} C[/tex]
W is the workdone whose value is obtained as = [tex]+2.20*10^{-6} J[/tex]
Substituting
[tex]\Delta V = \frac{W}{q}[/tex]
[tex]=\frac{2.2*10^{-6}}{4,20*10^{-9}}[/tex]
[tex]= 523.8 V[/tex]
Mathematically
[tex]\Delta V = Ed[/tex]
Where E is the electric field strength
d is the distanced moved and it is given as 6.00 cm = 0.06 m
Making E the subject of the formula
[tex]E = \frac{\Delta V}{d}[/tex]
[tex]E = \frac{523.8}{0.06} = 8730 \ N/C[/tex]
