Question

A sphere of radius R has charge Q. The electric field strength at distance$r>R$is$E_i$. What is the ratio $E_f/E_i$of the final to initial electric field strengths if

(a) Q is halved,

(b) R is halved, and

(c) r is halved (but is still$>R$)? Each part changes only one quantity; the other quantities have their initial values

Short answer

(a) The ratio of the final to initial electric field strengths, if Q is halved, is $1/2$.

(b) The ratio of the final to initial electric field strengths if radius R is halved is $1$.

(c) The ratio of the final to initial electric field strengths, if distance r is halved, is $4$.

Step-by-step solution

Given information (part a)

$$\begin{gathered} Charge=Q_i=Q \\ Charge{Q}_f=\frac{Q}{2} \\ {d}_i=r+R\cong r(Sincer>R) \\ {d}_f=r+R\cong r\left(Sincer>R\right) \end{gathered}$$

Explanation (part a)

$$\begin{gathered} sphere\mathrm{of}\mathrm{radius}{E}_i=\frac{Q_i}{4\pi {\epsilon }_0{d}_i^2} \\ Sphereofradius{E}_f=\frac{Q_f}{4\pi e_0{d}_f^2} \\ \Rightarrow \frac{E_f}{E_i}=\frac{\frac{{\mathrm{Q}}_{\mathrm{f}}}{4\pi e_0{\mathrm{r}}^2}}{\frac{\mathrm{Q}}{4\pi e_0{\mathrm{r}}^2}} \\ \Rightarrow \frac{E_f}{E_i}=\frac{Q}{2}\frac{1}{Q} \\ \Rightarrow \frac{E_f}{E_i}=\frac{1}{2} \end{gathered}$$

Given information (part b)

$$\begin{gathered} Sphereofradius{r}_i=R \\ Charge{Q}_i=Q \\ Charge{Q}_f=Q \\ {d}_i=\left(R+r\right) \\ {d}_f=\frac{R}{2}+\frac{R}{2}+r=\left(R+r\right) \end{gathered}$$

Explanation (part b)

$$\begin{gathered} sphere\mathrm{of}radius{E}_i=\frac{Q_i}{4\pi {\epsilon }_0{d}_i^2} \\ Sphereofradius{E}_f=\frac{Q_f}{4\pi e_0{d}_f^2} \\ \Rightarrow \frac{E_f}{E_i}=\frac{\frac{Q_f}{4\pi {\epsilon }_0(R+r{)}^2}}{\frac{Q_i}{4\pi {\tau }_0(R+r{)}^2}} \\ \Rightarrow \frac{E_f}{E_i}=\frac{Q}{1}\frac{1}{Q} \\ \Rightarrow \frac{E_f}{E_i}=\frac{1}{1} \\ \Rightarrow \frac{E_f}{E_i}=1 \end{gathered}$$

Given information (part c)

$$\begin{gathered} \text{Sphereofradius}{r}_i=R \\ Charge{Q}_i=Q \\ Sphereofradius{r}_f=R \\ Charge{Q}_f=Q \\ {d}_i=(\mathrm{R}+\mathrm{r})\cong r \\ {d}_f=\frac{r}{2}+R\cong \frac{r}{2} \end{gathered}$$

Explanation (part c)

$$\begin{gathered} sphere\mathrm{of}radius{E}_i=\frac{Q_i}{4\pi {\epsilon }_0{d}_i^2} \\ Sphereofradius{E}_f=\frac{Q_f}{4\pi e_0{d}_f^2} \\ \Rightarrow \frac{E_f}{E_i}=\frac{\frac{Q_f}{4\pi {ϵ}_0{\left(\frac{r}{2}\right)}^2}}{\frac{Q_i}{4\pi {ϵ}_0(r{)}^2}} \\ \Rightarrow \frac{E_f}{E_i}=\frac{Q}{1}\frac{4}{Q} \\ \Rightarrow \frac{E_f}{E_i}=4 \end{gathered}$$