mugenziwzn6

2023-02-22

A small metal sphere, carrying a net charge of q_1=−2μC, is held in a stationery position by insulating supports. A second small metal sphere, with a net charge of q_2=−8μC and mass 1.50 g is projected towards q_1. When the two spheres are 0.800m apart, q_2 is moving towards q_1 with speed 20 ms:−1 as shown in the figure. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. How close does q_2 get to q_1?

Emmalee Montes

Beginner2023-02-23Added 6 answers

Let i be the point when the charge q2 is 0.8 m.

Due to the same nature of the two charges, charge q2 will experience an electrostatic force of replusion from charge q1, which will cause charge q2's velocity to drop. Charge Q2 will eventually arrive at point a f, where it is the furthest away from charge Q1 and where its velocity is zero. Let r be the smallest distance between q1 and q2.

By total energy conservation

$\u27f9{E}_{i}={E}_{f}\phantom{\rule{0ex}{0ex}}P{E}_{i}+K{E}_{i}=P{E}_{i}+K{E}_{f}$

Velocity of charge q2 at point i = 20 ms^−1

∴ Kinetic energy of q2 at point i = 12m(20)2 J

Potential energy of charge q2 at point i will be kq1q2/0.8

Potential energy of charge q2 at point f will be kq1q2/r

Kinetic energy of q2 at point f = 0, vf=0

$\u27f9\frac{k{q}_{1}{q}_{2}}{0.8}+\frac{1}{2}(0.0015)(20{)}^{2}=\frac{k{q}_{1}{q}_{2}}{r}+0\phantom{\rule{0ex}{0ex}}\u27f90.48J=\frac{k{q}_{1}{q}_{2}}{r}\phantom{\rule{0ex}{0ex}}\u27f9r=\frac{9\times {10}^{9}\times 2\times {10}^{-6}\times 8\times {10}^{-6}}{0.48}\phantom{\rule{0ex}{0ex}}\therefore r=0.3m$

Due to the same nature of the two charges, charge q2 will experience an electrostatic force of replusion from charge q1, which will cause charge q2's velocity to drop. Charge Q2 will eventually arrive at point a f, where it is the furthest away from charge Q1 and where its velocity is zero. Let r be the smallest distance between q1 and q2.

By total energy conservation

$\u27f9{E}_{i}={E}_{f}\phantom{\rule{0ex}{0ex}}P{E}_{i}+K{E}_{i}=P{E}_{i}+K{E}_{f}$

Velocity of charge q2 at point i = 20 ms^−1

∴ Kinetic energy of q2 at point i = 12m(20)2 J

Potential energy of charge q2 at point i will be kq1q2/0.8

Potential energy of charge q2 at point f will be kq1q2/r

Kinetic energy of q2 at point f = 0, vf=0

$\u27f9\frac{k{q}_{1}{q}_{2}}{0.8}+\frac{1}{2}(0.0015)(20{)}^{2}=\frac{k{q}_{1}{q}_{2}}{r}+0\phantom{\rule{0ex}{0ex}}\u27f90.48J=\frac{k{q}_{1}{q}_{2}}{r}\phantom{\rule{0ex}{0ex}}\u27f9r=\frac{9\times {10}^{9}\times 2\times {10}^{-6}\times 8\times {10}^{-6}}{0.48}\phantom{\rule{0ex}{0ex}}\therefore r=0.3m$