Q. 1 a聽Charge q1is distance r聽from ... [FREE SOLUTION]

Chapter 25: Q. 1 (page 708)

$(a)$ Charge $q_{1}$ is distance $r$ from a positive point charge $Q$ . Charge $q_{2} = \frac{q_{1}}{3}$ is distance $2 r$ from $Q$ . What is the ratio $\frac{U_{1}}{U_{2}}$ of their potential energies due to their interactions with $Q$ ?
$(b)$ Charge $q_{1}$ is distance $s$ from the negative plate of a parallel-plate capacitor. Charge $q_{2} = \frac{q_{1}}{3}$ is distance $2 s$ from the negative plate. What is the ratio $\frac{U_{1}}{U_{2}}$ of their potential energies?

Short Answer

Expert verified

(a) The ratio of positive point charge, $\frac{U_{1}}{U_{2}} = 6 .$

(b) The ratio of negative plate,

\frac{U_{1}}{U_{2}} = \frac{3}{2} .

Step by step solution

Step: 1 Potential energy:

Energy that is stored 鈥� or preserved 鈥� in an item or material is referred to as potential energy. The stored energy is determined by the object's or substance's location, organisation, or condition.

Step: 2 Finding the ratio of positive point charges: (part a)

Interaction of potential energy $q_{1}$ with $Q$

$U_{1} = \frac{1}{4 蟺蔚_{0}} \frac{q_{1} Q}{r}$

Interaction of potential energy localid="1648738409728" $q_{2}$ with localid="1648738401056" $Q$

$U_{2} = \frac{1}{4 蟺蔚_{0}} \frac{q_{2}}{2 r} = \frac{1}{4 蟺蔚_{0}} \frac{q_{1}}{6 r}$

By dividing these equations

$\frac{U_{1}}{U_{2}} = \frac{1}{\frac{1}{6}} = 6 .$

Step: 2 Finding the ratio of negative plate: (part b)

A homogeneous electric field of magnitude $E$ is produced by the plate. The difference in potential between the point at distance $d$ from the plate and the plate is equal to

$螖 V = E d$

The potential energy is localid="1648738419356" $鈭� V$ times the charge applied to the spot.

$\begin{array}{l} U_{1} = q_{1} E s \\ U_{2} = q_{2} E 鈰� 2 s = \frac{q_{1}}{3} E 鈰� 2 s = \frac{2}{3} q_{1} E s . \end{array}$