Two protons of (charge +e) are held at rest separated by a distance 2d. One of the protons is released and flies away from the other due to the repulsive force between them. When the moving proton is a distance 3d from the other proton its kinetic energy is:

Question

(k_e*(3*e^2)/(2*d))
(k_e*(4*e^2)/(9*d))
(k_e*(e^2)/(4*d))
(k_e*(e^2)/(6*d))
(k_e*(2*e^2)/(3*d))

Answer 1

To find the kinetic energy of the moving proton when it is at a distance of $3d$ from the remaining proton, we can use conservation of energy.

Initially, the two protons are at rest and thus have only potential energy due to their electrostatic interaction. The initial potential energy $U_i$ when they are separated by a distance $2d$ is given by:

$$U_i=\frac{k_e{e}^2}{r}$$

where $k_e$ is Coulomb's constant and $r$ is the distance between the protons. Initially, the distance $r=2d$, so:

$$U_i=\frac{k_e{e}^2}{2d}$$

When one proton is released and moves to a distance of $3d$ from the stationary proton, the distance between them is now $r=3d$, and the potential energy $U_f$ at this point is:

$$U_f=\frac{k_e{e}^2}{3d}$$

According to the conservation of energy, the initial potential energy will be converted to the kinetic energy of the moving proton as well as the remaining potential energy when it is at a distance $3d$. The equation can be written as:

$$U_i=K+U_f$$

where $K$ is the kinetic energy of the moving proton at distance $3d$. Rearranging gives us:

$$K=U_i-U_f$$

Substituting the expressions for $U_i$ and $U_f$:

$$K=\frac{k_e{e}^2}{2d}-\frac{k_e{e}^2}{3d}$$

To perform the subtraction, we need a common denominator, which is $6d$:

$$K=(\frac{3k_e{e}^2}{6d}-\frac{2k_e{e}^2}{6d})$$

Now, simplifying this expression:

$$K=\frac{(3-2)k_e{e}^2}{6d}=\frac{k_e{e}^2}{6d}$$

Thus, the kinetic energy of the moving proton when it is at a distance of $3d$ from the other proton is:

$$\boxed{{\displaystyle \frac{k_e{e}^2}{6d}}}$$