Question

Two stationary positive point charges, charge 1 of magnitude 3.00 nC and charge 2 of magnitude 1.80 nC , are separated by a distance of 31.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed vfinal of the electron when it is 10.0 cm from charge 1?

Answer 1

Answer:

The velocity is $v= 5.9*10^6 m/s$

Explanation:

Generally the the potential at the middle of the two individual charges is mathematically represented as

               $V_1 = k [\frac{q_1}{\frac{d}{2} } + \frac{q_2}{\frac{d}{2} } ]$

Substituting  $9*10^9N$ for k (Coulomb constant ) , $3.95*10^{-9} C$ for $q_1$ , $1.80*10^{-9}C$ for $q_2$ and 31.0cm= $\frac{31}{100} = 0.31m$ for d

             $V_1 = (9*10^9) [\frac{3.95*10^{-9} ]}{\frac{0.31}{2} } + \frac{1.80*10^{-9}}{\frac{0.31}{2} } ]$

                 $= 333.87V$

The potential at 10cm from charge 1 is mathematically  evaluated as

                    $V_2 = k [\frac{q_1}{d_1} + \frac{q_2}{d_2} ]$

Substituting 10cm = $\frac{10}{100} = 0.10m$ for $d_1$ , (31-10)cm = $\frac{(31-10)}{100} = 0.21m$ for $d_2$ and the rest of the values

                  $V_2 = (9*10^9) [\frac{3.95*10^{-9}}{0.10} + \frac{1.80 *0^{-9}}{0.21} ]$

                      $=432.64V$

 According to the law of conservation of energy

   The difference in potential energy is equal to the kinetic energy

                    KE = ($V_2 -V_1$) $q$

Where is the charge on an electron

                    $\frac{1}{2} mv^2 = (V_2 - V_1)q$

substituting $9.1*10^{-31} kg$ for m (mass of electron) , $1.602 *10^{-19}C$ for q  (charge on an electron) , and making v the subject

                    $v = \sqrt{\frac{2(432.64-333.87) * 1.602*10^{-19}}{9.1*10^{-31}} }$

                        $v= 5.9*10^6 m/s$

                     

 

Answer 2

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