Respuesta :
Answer: U = -4.97*10^-17 J
Explanation:
Potential Energy of point charges,
U = kqq• / r, where
U = Potential Energy
q, q• = value of electric charges
k = 8.99*10^9 N.m²/C² constant of proportionality
r = distance between two charges
a) first electric potential due to electric field of first charge
q = 3*10^-9 C
q• = q(electron) = -1.602*10^-19 C
r = 0.5 * 31 cm = 15.5 cm = 0.155 m
U1 = kqq•/r
U1 = (8.99*10^9 * 3*10^-9 * -1.602*10^-19) / 0.155
U1 = -4.32*10^-18 / 0.155
U1 = -2.79*10^-17 J
Second electric potential due to electric field of second charge
U2 = kqq•/r
U2 = (8.99*10^9 * 1.8*10^-9 * -1.602*10^-19) / 0.155
U2 = -2.59*10^-18 / 0.155
U2 = -1.67*10^-17 J
U = U1 + U2
U = -2.79*10^-17 + -1.67*10^-17
U = -4.46*10^-17 J
b) first electric potential due to electric field of the first charge
q = 3*10^-9 C
q• = q(electron) = -1.602*10^-19 C
r = 10 cm = 0.1 m
U1 = kqq•/r
U1 = (8.99*10^9 * 3*10^-9 * -1.602*10^-19) / 0.1
U1 = -4.32*10^-18 / 0.1
U1 = -4.32*10^-17 J
Second electric potential to the electric field of second charge
q = 1.8*10^-9
q• = -1.602*10^-19 C
r = 50 - 10 = 40 cm = 0.4m
U2 = (8.99*10^9 * 1.8*10^-9 * -1.602*10^-19) / 0.4
U2 = -2.59*10^-18 / 0.4
U2 = -6.48*10^-18 J
U = U1 + U2
U = -4.32*10^-17 + -6.48*10^-18
U = -4.97*10^-17 J
Answer:
The velocity is [tex]v= 5.9*10^6 m/s[/tex]
Explanation:
Generally the the potential at the middle of the two individual charges is mathematically represented as
[tex]V_1 = k [\frac{q_1}{\frac{d}{2} } + \frac{q_2}{\frac{d}{2} } ][/tex]
Substituting [tex]9*10^9N[/tex] for k (Coulomb constant ) , [tex]3.95*10^{-9} C[/tex] for [tex]q_1[/tex] , [tex]1.80*10^{-9}C[/tex] for [tex]q_2[/tex] and 31.0cm= [tex]\frac{31}{100} = 0.31m[/tex] for d
[tex]V_1 = (9*10^9) [\frac{3.95*10^{-9} ]}{\frac{0.31}{2} } + \frac{1.80*10^{-9}}{\frac{0.31}{2} } ][/tex]
[tex]= 333.87V[/tex]
The potential at 10cm from charge 1 is mathematically evaluated as
[tex]V_2 = k [\frac{q_1}{d_1} + \frac{q_2}{d_2} ][/tex]
Substituting 10cm = [tex]\frac{10}{100} = 0.10m[/tex] for [tex]d_1[/tex] , (31-10)cm = [tex]\frac{(31-10)}{100} = 0.21m[/tex] for [tex]d_2[/tex] and the rest of the values
[tex]V_2 = (9*10^9) [\frac{3.95*10^{-9}}{0.10} + \frac{1.80 *0^{-9}}{0.21} ][/tex]
[tex]=432.64V[/tex]
According to the law of conservation of energy
The difference in potential energy is equal to the kinetic energy
KE = ([tex]V_2 -V_1[/tex]) [tex]q[/tex]
Where is the charge on an electron
[tex]\frac{1}{2} mv^2 = (V_2 - V_1)q[/tex]
substituting [tex]9.1*10^{-31} kg[/tex] for m (mass of electron) , [tex]1.602 *10^{-19}C[/tex] for q (charge on an electron) , and making v the subject
[tex]v = \sqrt{\frac{2(432.64-333.87) * 1.602*10^{-19}}{9.1*10^{-31}} }[/tex]
[tex]v= 5.9*10^6 m/s[/tex]
