Question

Two stationary positive point charges, charge 1 of magnitude 3.05 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 41.0 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

To solve this problem we will apply the concepts related to voltage as a dependent expression of the distance of the bodies, the Coulomb constant and the load of the bodies. In turn, we will apply the concepts related to energy conservation for which we can find the speed of this

$V = \frac{kq}{r}$

Here,

k = Coulomb's constant

q = Charge

r = Distance to the center point between the charge

From each object the potential will be

$V_1 = \frac{kq_1}{r_1}+\frac{kq_2}{r_2}$

Replacing the values we have that

$V_1 = \frac{(9*10^9)(3.05*10^{-9})}{0.41/2}+\frac{(9*10^9)(1.85*10^{-9})}{0.41/2}$

$V_1 = 215.12V$

Now the potential two is when there is a difference at the distance of 0.1 from the second charge and the first charge is 0.1 from the other charge, then,

$V_1 = \frac{(9*10^9)(3.05*10^{-9})}{0.1}+\frac{(9*10^9)(1.85*10^{-9})}{0.41-0.1}$

$V_2 = 328.2V$

Applying the energy conservation equations we will have that the kinetic energy is equal to the electric energy, that is to say

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

Here

m = mass

v = Velocity

q = Charge

V = Voltage

Rearranging to find the velocity

$v = \sqrt{ \frac{2q(V_2-V_1)}{m}}$

Replacing,

$v = \sqrt{ \frac{-2(1.6*10^{-19})(328.2-215.12)}{9.11*10^{-3}}}$

$v = 6.3*10^6m/s$

Therefore the speed final velocity of the electron when it is 10.0 cm from charge 1 is $6.3*10^6m/s$

Answer 2

The final speed of electron is $6.3 \times 10^{6} \;\rm m/s$.

Given data:

The magnitude of charge 1 is, q = 3.05 nC.

The magnitude of charge 2 is, q' = 1.85 nC.

The distance of separation is, d = 41.0 cm = 0.41 m.

The distance between the electron and charge 1 is, d' = 10.0 cm = 0.1 m.

Apply the concepts related to voltage as a dependent expression of the distance of the bodies, the Coulomb constant and the load of the bodies. In turn, we will apply the concepts related to energy conservation.

Net potential  between the charged entities and electrons is,

$V = \dfrac{kq}{d/2}+\dfrac{kq'}{d/2}$

Here, k is the Coulomb's constant.

Solving as,

$V = \dfrac{9 \times 10^{9} \times 3.05 \times 10^{-9}}{0.41/2}+\dfrac{9 \times 10^{9} \times 1.85 \times 10^{-9}}{0.41/2}\\\\V = 215.12 \;\rm V$

Now the potential difference when there is a difference at the distance of 0.1 m from the second charge and the first charge is 0.1 from the other charge as,

$V' = \dfrac{kq}{d'}+\dfrac{kq'}{d'}\\\\V' = \dfrac{9 \times 10^{9} \times 3.05 \times 10^{-9}}{0.1}+\dfrac{9 \times 10^{9} \times 1.85 \times 10^{-9}}{0.1}\\\\V' =328.2 \;\rm V$

Now, as per the energy conservation equations we will have that the kinetic energy is equal to the electric energy,

$KE = q(V'-V)\\\\\dfrac{1}{2}mv^{2} = q(V'-V)$

Here, m is the mass of electron, v is the final speed of electron and q is the charge on electron.

Solving as,

$\dfrac{1}{2} \times 9.1 \times 10^{-31} \times v^{2} = 1.6 \times 10^{-19} \times (328.2 -215.12)\\\\v =\sqrt{\dfrac{2 \times 1.6 \times 10^{-19} \times (328.2-215.12)}{ 9.1 \times 10^{-31}}}\\\\v = 6.3 \times 10^{6} \;\rm m/s$

Thus, we conclude that the final speed of electron is $6.3 \times 10^{6} \;\rm m/s$.

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