Two stationary point charges +3 nC and + 2nC are separated bya distance of 50cm

Given:
Charge 1, Q1 = +3nC
Charge 2, Q2 = +2nC
Distance between the charges, d = 50 cm
Distance of the electron from the +3nC charge, x = 10 cm
The potential energy of the electron at the midpoint is given by the equation:
$P{E}_{\text{initial}}=\frac{k·|Q1·Q2|}{d}$
where k is the electrostatic constant, which is approximately $8.99\times {10}^9{\text{Nm}}^2/{\text{C}}^2$.
The final kinetic energy of the electron when it is 10 cm from the +3nC charge is given by:
$K{E}_{\text{final}}=\frac{1}{2}·m·v^2$
where m is the mass of the electron, which is approximately $9.11\times {10}^{-31}$ kg, and v is the speed of the electron.
Since energy is conserved, we can equate the initial potential energy to the final kinetic energy:
$\frac{k·|Q1·Q2|}{d}=\frac{1}{2}·m·v^2$
Now we can plug in the given values:
$k=8.99\times {10}^9{\text{Nm}}^2/{\text{C}}^2$
$Q1=+3\times {10}^{-9}$ C
$Q2=+2\times {10}^{-9}$ C
$d=50$ cm
$m=9.11\times {10}^{-31}$ kg
$x=10$ cm
Substituting these values into the equation, we can solve for v:
$\frac{(8.99\times {10}^9{\text{Nm}}^2/{\text{C}}^2)·|(3\times {10}^{-9}\text{C})·(2\times {10}^{-9}\text{C})|}{(50\times {10}^{-2}\text{m})}=\frac{1}{2}·(9.11\times {10}^{-31}\text{kg})·v^2$
Simplifying the equation:
$\frac{8.99\times 3\times 2\times {10}^{-18}}{50\times {10}^{-2}}=\frac{1}{2}·9.11\times v^2$
$1.0788\times {10}^{-15}=4.5555\times {10}^{-31}·v^2$
Now, solve for v:
$v^2=\frac{1.0788\times {10}^{-15}}{4.5555\times {10}^{-31}}$
$v^2=2.3663\times {10}^{15}$
$v=\sqrt{2.3663\times {10}^{15}}$
$v\approx 4.865\times {10}^7$ m/s
Therefore, the speed of the electron when it is 10 cm from the +3nC charge is approximately $4.865\times {10}^7$ m/s.

Answer:
$7.89\times {10}^{17}$ m/s
Explanation:
Let's denote the charge of the electron as e and the distance between the charges as d. The given values are:
Charge of +3nC charge (Q1) = +3nC = $3\times {10}^{-9}$ C
Charge of +2nC charge (Q2) = +2nC = $2\times {10}^{-9}$ C
Distance between the charges (d) = 50 cm = 0.5 m
Distance from the +3nC charge to the electron (r) = 10 cm = 0.1 m
The electric potential energy (PE) of the electron at the midpoint is given by:
$PE=\frac{Q1·Qe}{4\pi {ϵ}_0·d/2}+\frac{Q2·Qe}{4\pi {ϵ}_0·d/2}$
where Qe is the charge of the electron and ${ϵ}_0$ is the permittivity of free space.
The initial potential energy (PEi) of the electron at the midpoint is:
$PEi=\frac{Q1·Qe}{4\pi {ϵ}_0·d/2}+\frac{Q2·Qe}{4\pi {ϵ}_0·d/2}$
The final kinetic energy (KEf) of the electron when it is 10 cm away from the +3nC charge is:
$KEf=\frac{1}{2}m·v^2$
where m is the mass of the electron and v is its velocity.
Since the electron is released from rest, its initial kinetic energy (KEi) is zero.
By the conservation of energy principle, we have:
$PEi=KEf$
Substituting the values into the equation and solving for v:
$\frac{Q1·Qe}{4\pi {ϵ}_0·d/2}+\frac{Q2·Qe}{4\pi {ϵ}_0·d/2}=\frac{1}{2}m·v^2$
Now we can substitute the known values:
$\frac{(3\times {10}^{-9}C)·(-1.6\times {10}^{-19}C)}{4\pi {ϵ}_0·(0.5/2)m}+\frac{(2\times {10}^{-9}C)·(-1.6\times {10}^{-19}C)}{4\pi {ϵ}_0·(0.5/2)m}=\frac{1}{2}(9.1\times {10}^{-31}kg)·v^2$
Simplifying the equation and solving for v:
$5.4\times {10}^6-2.56\times {10}^6=4.55\times {10}^{-31}·v^2$
$2.84\times {10}^6=4.55\times {10}^{-31}·v^2$
$v^2=\frac{2.84\times {10}^6}{4.55\times {10}^{-31}}$
$v=\sqrt{\frac{2.84\times {10}^6}{4.55\times {10}^{-31}}}$
$v=\sqrt{6.24\times {10}^{36}}$
$v\approx 7.89\times {10}^{17}$ m/s
Therefore, the speed of the electron when it is 10 cm away from the +3nC charge is approximately $7.89\times {10}^{17}$ m/s.

Step 1:
The electric potential energy between two point charges can be calculated using the formula:
$U={\displaystyle \frac{k·q_1·q_2}{r}}$,
where $U$ is the electric potential energy, $k$ is the electrostatic constant ($9\times {10}^9\text{N}·{\text{m}}^2/{\text{C}}^2$), $q_1$ and $q_2$ are the charges, and $r$ is the distance between the charges.
At the midpoint between the two charges, the electric potential energy of the electron is zero since it is equidistant from both charges.
As the electron moves towards the +3nC charge, it gains potential energy. At a distance of 10cm from the +3nC charge, the potential energy gained by the electron can be calculated using the formula mentioned above. This potential energy will be converted into kinetic energy, given by the equation:
$K={\displaystyle \frac{1}{2}}m{v}^2$,
where $K$ is the kinetic energy, $m$ is the mass of the electron ($9.11\times {10}^{-31}\text{kg}$), and $v$ is the velocity (speed) of the electron.
Since the total mechanical energy (the sum of kinetic and potential energies) is conserved, we can equate the potential energy gained by the electron to its kinetic energy at a distance of 10cm from the +3nC charge:
${\displaystyle \frac{k·q_1·q_{\text{electron}}}{r}}={\displaystyle \frac{1}{2}}m{v}^2$,
where $q_{\text{electron}}$ represents the charge of the electron (-1.6nC).
Step 2:
Now we can solve this equation for $v$, the velocity (speed) of the electron.
Remember to substitute the appropriate values for $k$, $q_1$, $q_{\text{electron}}$, $r$, and $m$ before solving the equation.
${\displaystyle \frac{(9\times {10}^9\text{N}·{\text{m}}^2/{\text{C}}^2)·(3\times {10}^{-9}\text{C})·(-1.6\times {10}^{-9}\text{C})}{0.1\text{m}}}={\displaystyle \frac{1}{2}}·(9.11\times {10}^{-31}\text{kg})·v^2$
Now, we can solve for $v$:
$5.4\times {10}^{-5}\text{N}·\text{m}=(4.5555\times {10}^{-31}\text{kg})·v^2$
Dividing both sides by $(4.5555\times {10}^{-31}\text{kg})$:
$v^2={\displaystyle \frac{5.4\times {10}^{-5}\text{N}·\text{m}}{4.5555\times {10}^{-31}\text{kg}}}$
$v^2\approx 1.184\times {10}^6{\text{m}}^2/{\text{s}}^2$
Taking the square root of both sides:
$v\approx 1.088\times {10}^3\text{m/s}$
Therefore, the speed of the electron when it is 10cm from the +3nC charge is approximately $1.088\times {10}^3\text{m/s}$.