Question

A small segment of wire in FIGURE $Q23.4$contains $10nC$of charge.

a. The segment is shrunk to one-third of its original length. What is the ratio of ${\lambda }_f/{\lambda }_i$, where ${\lambda }_{i}and{\lambda }_f$ are the initial and final linear charge densities?

b. A proton is very far from the wire. What is the ratio Ff /Fi of the electric force on the proton after the segment is shrunk to the force before the segment was shrunk?

c. Suppose the original segment of wire is stretched to $10$ times its original length. How much charge must be added to the wire to keep the linear charge density unchanged?

Short answer

(a) The ratio of final to initial linear charge densities when the length shrunk to one-third of its original length$=3$

(b) The ratio of final to initial linear forces when the length shrunk to one-third of its original length$=3$.

(c) The amount of final charge required to keep linear charge densities constant when its final length is stretched 10 times of initial length$=100nC$

Step-by-step solution

Given information (part a)

$$\begin{gathered} \text{Case}\left(1\right): \\ \text{length}{l}_i=l \\ \text{Charge}{Q}_i=10nC \\ \text{Case}\left(2\right): \\ \text{length}{l}_f=\frac{l}{3} \\ \text{Charge}{Q}_f=10nC \\ \text{Ratioofchargedensities}=\frac{{\lambda }_f}{{\lambda }_i} \end{gathered}$$

Explanation (part a)

$$\begin{gathered} \text{Linear charge density}\lambda =\frac{\text{charge}}{\text{length}} \\ \text{RatioofLinearchargedensities}=\frac{{\lambda }_f}{{\lambda }_i}=\frac{\frac{Q_f}{l_f}}{\frac{Q_i}{l_i}}\dots \dots \left(1\right) \\ \Rightarrow \frac{{\lambda }_f}{{\lambda }_i}=\frac{l_i}{l_f}\cdots \left(2\right)\dots \dots \left(2\right) \end{gathered}$$

Substituting the values of given data in $\text{eqn}\left(2\right)i.e.\text{,}\frac{{\lambda }_f}{{\lambda }_i}=\frac{l_i}{l_f}$

$$\begin{gathered} \Rightarrow \frac{{\lambda }_f}{{\lambda }_i}=\frac{l}{\frac{1}{3}} \\ \Rightarrow \frac{{\lambda }_f}{{\lambda }_i}=3 \end{gathered}$$

Given information (part b)

$$\begin{gathered} \text{Case}\left(1\right): \\ \text{length}{l}_i=l \\ \text{Charge}{Q}_i=10nC \\ Chargeofproton{Q}_p=Q\left(say\right) \\ \text{Case}\left(2\right): \\ \text{length}{l}_f=\frac{l}{3} \\ \text{Charge}{Q}_f=10nC \\ Chargeofproton{Q}_p=Q\left(say\right) \\ Ratioofforces=\frac{F_f}{F_i} \end{gathered}$$

Explanation (part b)

Substituting the values of given data in$$\begin{gathered} eqn\left(2\right)\text{i.e.,}\frac{{\lambda }_f}{{\lambda }_i}=\frac{l_i}{l_f} \\ Where,Linear\mathrm{charge}\mathrm{density}\lambda =\frac{\text{charge}}{\text{length}}\dots \dots \left(2\right) \\ \Rightarrow \frac{F_f}{F_i}=\frac{l}{\frac{l}{3}} \\ \Rightarrow \frac{F_f}{F_i}=3 \end{gathered}$$

Given information (part c)

$$\begin{gathered} \text{Case}\left(1\right): \\ \text{length}{l}_i=l \\ \text{Charge}{Q}_i=10nC \\ \text{Case}\left(2\right): \\ \text{length}{l}_f=10l \\ \text{Ratioofchargedensities}=\frac{{\lambda }_f}{{\lambda }_i}=1 \end{gathered}$$

Explanation (part c)

Substituting the values of given data in $eqn(2$)

$$\begin{gathered} Linearchargedensity\lambda =\frac{\text{charge}}{\text{length}} \\ Ratiooflinearchargedensities=\frac{{\lambda }_f}{{\lambda }_i}=1=\frac{\frac{{\ell }_f}{l_f}}{\frac{Q_i}{l_i}}\dots \dots \left(1\right) \\ \Rightarrow Q_f=\frac{l_2\times Q_i}{l_1}\dots \dots \left(2\right) \\ {Q}_f=\frac{10l\times 10\mathrm{nC}}{l} \\ \Rightarrow Q_f=100\mathrm{nC} \end{gathered}$$