Q45E - University Physics with Modern Physics

Question

  A metal sphere with radius $r_a=1.20 cm$ is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius $r_b=9.60 cm$. Charge $\mathbf{+}\mathbf{q}$ is put on the inner sphere and charge -q on the outer spherical shell. The magnitude of $\mathbf{q}$ is chosen to make the potential difference between the spheres $\mathbf{500}\mathbf{V}$ with the inner sphere at higher potential. (a) Use the result of Exercise $\mathbf{23}\mathbf{.}\mathbf{41}\left(b\right)$to calculate $\mathbf{q}$. $\left(b\right)$ With the help of the result of Exercise $\mathbf{23}\mathbf{.}\mathbf{41}\left(a\right)$,sketch the equipotential surfaces that correspond to $\mathbf{500}\mathbf{,}\mathbf{ }\mathbf{400}\mathbf{,}\mathbf{ }\mathbf{300}\mathbf{,}\mathbf{ }\mathbf{200}\mathbf{,}\mathbf{ }\mathbf{100}\mathbf{,}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{0}\mathbf{ }\mathbf{V}\mathbf{.}$$\left(c\right)$ In your sketch, show the electric field lines. Are the electric field lines and equipotential surfaces mutually perpendicular? Are the equipotential surfaces closer together when the magnitude of $\overrightarrow{E}$ is largest?

Step-by-Step Solution

Answer

(a)The value of $\mathbf{q}$ from Exercise $\mathbf{23}\mathbf{.}\mathbf{41}\left(b\right)$ is $7.6*{10}^{-10}C$.

(b)The picture is as shown below.

(c) Yes. The electric field lines and equipotential surfaces are mutually perpendicular. Yes. The equipotential surfaces seem to be closer together wherever the magnitude of the electric field is largest.

Step 1: Finding the value of q.

The potential difference between two concentric charged spheres of radii $r_a$and $r_b$ is expressed as $V=kq\left[\frac{1}{r_a}-\frac{1}{r_b}\right]$, where $q$is the magnitude of the charge and k is the constant $\frac{1}{4p{e}_o}=k=9*{10}^{9}N{m}^2/C^2$

 

Thus the expression for the potential difference between any surface of radius $r$ with respect to a known radius surface $r_a$, is $V=\frac{kq}{r_a}-\frac{kq}{r}$.

Step 2: Calculation of the value of q.

       (a) Solving for $q$ in $V=kq\left[\frac{1}{r_a}-\frac{1}{r_b}\right]$, implies $q=\frac{V}{k\left[\frac{1}{r_a}-\frac{1}{r_b}\right]}$.

Substituting the values of $r_a=1.2cm=1.2*{10}^{-2}m$ and $r_b=9.6cm=9.6*{10}^{-2}m$, and the potential difference V given as  

 and in the expression $q=\frac{V}{k\left[{\displaystyle \frac{1}{r_a}}-{\displaystyle \frac{1}{r_b}}\right]}$,

                                $$\begin{gathered} q=\frac{500}{\left(9*{10}^9\right)\left[{\displaystyle \frac{1}{\left(1.2*{10}^{-2}\right)}}-{\displaystyle \frac{1}{\left(9.6*{10}^{-2}\right)}}\right]} \\ q=\frac{500}{\left(9*{10}^9\right)\left[{\displaystyle 83.334-10.417}\right]} \\ q=\frac{500}{\left(9*{10}^9\right)\left[{\displaystyle 72.917}\right]} \\ q=7.619*{10}^{-10}C \end{gathered}$$

 

Thus, the charge contained in both the spheres are of magnitude $q=7.6*{10}^{-10}C$, as $+7.6*{10}^{-10}C$ in the inner shell and $-7.6*{10}^{-10}C$in the outer shell.

Step 3: Skteching the equipotential surfaces.

The result from exercise $\mathbf{23}\mathbf{.}\mathbf{41}\left(b\right)$ used are, the expressions for V as the distance varies. At $r<r_a$, the potential difference is constant as due to charge at $r_a$ itself, denoted by $V_a$ as there is no enclosed charge within $r_a$. At $r_a<r<r_b$, the enclosed charge is q, so $V_{rb}=kq\left[\frac{1}{r}-\frac{1}{r_b}\right]$. At $r>r_b$, net enclosed charge is zero, $V\left(r\right)=0$.  Generally, in the annular section $V_{ar}=\frac{kq}{r_a}-\frac{kq}{r}$ works well. 

 

Find the distance of separation between each concentric equipotential surfaces by solving for r, in $V_{ar}=\frac{kq}{r_a}-\frac{kq}{r}$, giving  $\frac{1}{r}=-\frac{V_a-V_r}{kq}+\frac{1}{r_a}$. 

The equipotential surfaces need to be drawn for V=500,400,300,200,100,0V.

Given $V_{ab}=500V$

For $V=500V\frac{1}{r}=-\frac{V_a-V_r}{kq}+\frac{1}{r_a}$becomes,

                            $$\begin{gathered} \frac{1}{r}=-\frac{500 -500}{\left(9*{10}^9\right)\left(7.619*{10}^{-10}\right)}+\frac{1}{\left(1.2*{10}^{-2}\right)} \\ \frac{1}{r}=\frac{1}{\left(1.2*{10}^{-2}\right)} \\ r=1.2*{10}^{-2}=0.012m \end{gathered}$$

which is the distance from centre. Then the distance with respect to the inner sphere of radius $r_a=0.012m$ is, $d=r-r_a$, that  gives $d=0.012-0.012=0$. Thus, the 500V equipotential surface lies on the inner shell itself.

For $V=400V\frac{1}{r}=-\frac{V_a-V_r}{kq}+\frac{1}{r_a}$becomes,

$$\begin{gathered} \frac{1}{r}=-\frac{500-400}{(9*{10}^9)(7.619*{10}^{-10})}+\frac{1}{(1.2*{10}^{-2})} \\ \frac{1}{r}=-\frac{100}{6.8571}+\frac{1}{\left(0.012\right)} \\ \frac{1}{r}=-14.583+83.334 \\ r=0.0145m \end{gathered}$$

Now, $d=r-r_a$ is$d=0.0145-0.012=0.0025m$. Thus, the 400V equipotential surface lies at a distance of $0.0025m$ from the  inner shell.

 

For $V=300V\frac{1}{r}=-\frac{V_a-V_r}{kq}+\frac{1}{r_a}$becomes,

$$\begin{gathered} \frac{1}{r}=\frac{500-300}{\left(9*{10}^9\right)\left(7.619*{10}^{-10}\right)}+\frac{1}{\left(1.2*{10}^{-2}\right)} \\ \frac{1}{r}=-\frac{200}{6.8571}+\frac{1}{\left(0.012\right)} \\ \frac{1}{r}=-29.1668+83.334 \\ r=0.01846m \end{gathered}$$

Now, $d=r-r_a$ is $d=0.01846-0.012=0.00646m$. Thus, the $300V$ equipotential surface lies at a distance of $0.00646m$ from the  inner shell. That is the separation between $400V$and $300V$ shells is thus, $d_{400-300}=0.0025: 0.00646=0.00396m$, which is at a larger separation than that between 500V to 400V surfaces.

For $V=200V\frac{1}{r}=-\frac{V_a-V_r}{kq}+\frac{1}{r_a}$becomes,

$$\begin{gathered} \frac{1}{r}=-\frac{500-200}{(9*{10}^9)(7.619*{10}^{-10})}+\frac{1}{(1.2*{10}^{-2})} \\ \frac{1}{r}=-\frac{300}{6.8571}+\frac{1}{\left(0.012\right)} \\ \frac{1}{r}=-43.7503+83.334 \\ r=0.02526m \end{gathered}$$

Now, $d=r-r_a$ is $d=0.02526-0.012=0.01326m$. Thus, the $200V$equipotential surface lies at a distance of $0.01326m$ from the  inner shell. That is the separation between $200V$ and $300V$ shells is thus, $d_{200-300}=0.00646~0.01326=0.0068m$, which is at a larger separation than that between 300V to 400V surfaces.

 

For $V=100V,\frac{1}{r}=-\frac{V_a-V_r}{kq}+\frac{1}{r_a}$, becomes,

$$\begin{gathered} \frac{1}{r}=-\frac{500-100}{(9*{10}^9)(7.619*{10}^{-10})}+\frac{1}{(1.2*{10}^{-2})} \\ \frac{1}{r}=-\frac{400}{6.8571}+\frac{1}{\left(0.012\right)} \\ \frac{1}{r}=-58.3336+83.334 \\ r=0.03999m \end{gathered}$$

Now, $d=r-r_a$ is $d=0.03999-0.012=0.02799m$. Thus, the $100V$ equipotential surface lies at a distance of $0.02799m$ from the  inner shell. That is the separation between $200V$ and $300V$ shells is thus, $d_{200-300}=0.01326: 0.02799=0.01473m$, which is at a larger separation than that between 200V to 300V surfaces.

For $V=0V$, $\frac{1}{r}=-\frac{V_a-V_r}{kq}+\frac{1}{r_a}$becomes,

$$\begin{gathered} \frac{1}{r}=-\frac{500}{(9*{10}^9)(7.619*{10}^{-10})}+\frac{1}{(1.2*{10}^{-2})} \\ \frac{1}{r}=-\frac{500}{6.8571}+\frac{1}{\left(0.012\right)} \\ \frac{1}{r}=-72.9171+83.334 \\ r=0.09599m \end{gathered}$$

Now, $d=r-r_a$ is $d=0.09599-0.012=0.08399m$. Thus, the  $0V$equipotential surface lies at a distance of $0.08399m$ from the  inner shell. That is the separation between $100V$ and 0V shells is thus, $d_{100-0}=0.08399~0.02799=0.056m$, which is at a larger separation than that between 200V to 100V surfaces.

Step 4: Relationship between electric field value and equipotential surfaces separation.

(c) Yes, the electric field lines are perpendicular to the equipotential surfaces.

As the potential difference decreases, the distance of separation seems to increase or vice versa. The electric field is given by the negative gradient of potential, $\left[E=-\frac{d[V_{final}-V_{initial}]}{dx}=\frac{V_{initial-}{V}_{final}}{dx}\right]$ the electric field is proportional to the magnitude of potential difference. Thus, greater the potential difference, greater is the electric field intensity, then, lesser is the distance between equipotential surfaces.