A proton is moving away from an electron. Fin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The electrostatic potential energy $U$ of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $U = \frac{k q_1 q

Vedclass · Accepted Answer

will increase

Vedclass · Answer

(B) The electrostatic potential energy $U$ of a system of two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by $U = \frac{k q_1 q_2}{r}$,where $k$ is Coulomb's constant.For a proton $(+e)$ and an electron $(-e)$,the potential energy is $U = \frac{k(e)(-e)}{r} = -\frac{k e^2}{r}$.As the proton moves away from the electron,the separation distance $r$ increases.Since $r$ is in the denominator and the potential energy is negative,as $r$ increases,the value of $-\frac{k e^2}{r}$ becomes less negative (i.e.,it approaches zero from the negative side).Therefore,the potential energy of the system increases.

$A$ proton is moving away from an electron. Find the change in the potential energy of the system.

Similar Questions

Two charges $5 \text{ nC}$ and $-2 \text{ nC}$ are placed at points $(5 \text{ cm}, 0, 0)$ and $(23 \text{ cm}, 0, 0)$ in a region of space where there is no other external field. The electrostatic potential energy of this charge system is

When a dipole placed parallel to electric field is rotated through $\pi^c$, the work done is $W$. The work done in rotating the dipole through $(\frac{\pi}{3})^c$ is $(\cos 0^\circ = 1, \cos 60^\circ = \frac{1}{2}, \cos 180^\circ = -1)$

Nine point charges are placed on a cube as shown in the figure. The charge $q$ is placed at the body centre whereas all other charges are at the vertices. The electrostatic potential energy of the system will be

When one electron is moved towards another electron,the electric potential energy of the system:

An electric dipole with dipole moment $\vec p$ is placed in a uniform electric field $\vec E$. The work done in rotating the dipole by $90^\circ$ from its initial equilibrium position is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module