Question

A capacitor consists of two charged disks of radius$\mathit{R}$ separated by a distance$\mathit{s}$, where$\mathit{R}\mathbf{>}\mathbf{>}\mathit{s}$. The magnitude of the charge on each disk is Q. Consider points A, B, C, and D inside the capacitor, as shown in Figure 16.88. (a) Show that$\mathit{\Delta }\mathit{V}\mathbf{=}{\mathit{V}}_{\mathbf{C}}\mathbf{-}{\mathit{V}}_{\mathbf{A}}$is the same for these paths by evaluating ∆V along each path: (1) Path 1:A = B = C, (2) Path 2:$\mathbf{A}\mathbf{=}\mathbf{C}$, (3) Path 3:$\mathit{A}\mathbf{=}\mathit{D}\mathbf{=}\mathit{B}\mathbf{=}\mathit{C}$. (b) If,$\mathit{Q}\mathbf{=}\mathbf{43}\mathbf{\text{\hspace{0.17em}μC}}$,$\mathit{R}\mathbf{=}\mathbf{4}\mathbf{\text{\hspace{0.17em}m}}$,${\mathit{s}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.5}\mathbf{\text{\hspace{0.17em}mm}}$ and${\mathit{s}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.7}\mathbf{\text{\hspace{0.17em}mm}}$, what is the value of$\mathit{\Delta }\mathit{V}\mathbf{=}{\mathit{V}}_{\mathbf{C}}\mathbf{-}{\mathit{V}}_{\mathbf{A}}$? (c) Choose two different paths from point A back to point A again, and show that$\mathbf{∆}\mathit{V}\mathbf{=}\mathbf{0}$for a round trip along both of these paths.

Short answer

(a) It is proved that potential difference $V_C-V_A$is same for every path and its value is$\frac{Q{s}_1}{\pi {\epsilon }_0{R}^2}$ .

(b) The value of potential difference is $144.92\text{\hspace{0.17em}V}$.

(c) It is proved that the potential difference in any path forming a close loop is zero.

Step-by-step solution

Identification of given data

The given data is listed below,

• The radius of each disk is given by,$R=4\text{\hspace{0.17em}m}$
• The charge on the disks is,$\mathrm{Q}=43\text{\hspace{0.17em}μC}$
• The disk separation is,$s$
• The separation between point A and B is,$s_1=1.5\text{\hspace{0.17em}mm}\times \left(\frac{1\text{\hspace{0.17em}m}}{1000\text{\hspace{0.17em}mm}}\right)=1.5\times {10}^{-3}\text{\hspace{0.17em}m}$
• The separation between point B and C is,$s_2=0.7\text{\hspace{0.17em}mm}\times \left(\frac{1\text{\hspace{0.17em}m}}{1000\text{\hspace{0.17em}mm}}\right)=0.7\times {10}^{-3}\text{\hspace{0.17em}m}$

Significance of potential difference.

When a test charge is placed within the influence area of a source charge, it acquires some electric potential. If the test charge moves between two points within the influence area of the source charge, there will be differences in the potential between those two points, which is referred to as potential difference.

(a) Determination of the potential difference VC−VA  for three paths

1. Path $A\to B\to C$.

The constant electric field in $x$-direction is given by,

$\overrightarrow{E}=\frac{Q}{A{\epsilon }_0}\widehat{x}$

Here, Q is the charge on the capacitor, A is the area of the capacitor, ${\epsilon }_0$is the dielectric constant, and$\widehat{x}$is the unit vector in the $x$-direction.

Taking path from A to B and then to C. as the electric field only has a component on x axis, the path from B to C does not contribute, so the potential difference $V_C-V_A$ can be given by,

$\begin{array}{c}{V}_C-V_A=V_B-V_A\\ =-E_x(x_C-x_A)\\ =-\frac{Q}{A{\epsilon }_0}(-s_1)\\ =\frac{Q{s}_1}{\pi {\epsilon }_0{R}^2}\end{array}$

Here, R is the radius of charged disk, $x_C-x_A$is the distance between point A and C.

2. Diagonal path from A to C.

The potential difference $V_C-V_A$is given by,

$\begin{array}{c}{V}_C-V_A=-E_x\Delta x-E_y\Delta y\\ =-E_x(x_C-x_A)\\ =-\frac{Q}{A{\epsilon }_0}(-s_1)\\ =\frac{Q{s}_1}{\pi {\epsilon }_0{R}^2}\end{array}$

Taking a path from A to C diagonally both x component and y component will be included in the potential difference since there is no electric field in y-direction y component of electric field will become zero and only x component of electric field will contribute in potential difference

(3) Path$A\to D\to B\to C$

The path will start from A to D, then from D to B and finally from B to C. the field only has a component in x- axis, so, the paths from A to D and B to C do not contribute in the potential difference $V_C-V_A$, which can be given by,

$\begin{array}{c}{V}_C-V_A=V_B-V_D\\ =-E_x\Delta x-E_y\Delta y\\ =-E_x(x_C-x_A)\\ =-\frac{Q}{A{\epsilon }_0}(-s_1)\\ =\frac{Q{s}_1}{\pi {\epsilon }_0{R}^2}\end{array}$

Here,$V_C-V_A$ is the potential difference between point A and C,$V_B-V_D$is the potential difference between point B and D, $x_C-x_A$is distance between point A and C denoted by $-s_1$

Thus, it is proved that potential difference $V_C-V_A$is same for every path and its value is $\frac{Q{s}_1}{\pi {\epsilon }_0{R}^2}$.

(b) Calculation of the final numerical value of the potential difference VC−VA .

The potential difference between the point A and C is given by,

$V_C-V_A=\frac{Q{s}_1}{\pi {\epsilon }_0{R}^2}$

Here,Q is the charge on the capacitor, R is the radius of the disk, and $s_1$is the separation between point A and C.

Substitute all the given values in the above equation.

$\begin{array}{c}{\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{A}}=\frac{43\times {10}^{-6}\text{\hspace{0.17em}C}(1.5\times {10}^{-3}\text{\hspace{0.17em}m})}{\mathrm{\pi }(8.85\times {10}^{-12}{\text{\hspace{0.17em}C}}^2\text{/N}\cdot {\text{m}}^2){\left(4\text{\hspace{0.17em}m}\right)}^2}\\ =144.92\text{\hspace{0.17em}N}\cdot \text{m/C}\times \left(\frac{1\text{\hspace{0.17em}V}}{1\text{\hspace{0.17em}N}\cdot \text{m/C}}\right)\\ =144.92\text{\hspace{0.17em}V}\end{array}$

Thus, the value of potential difference is $144.92\text{\hspace{0.17em}V}$.

(c) Determination of potential difference between point A to A will always be zero irrespective of any path.

The potential difference along a closed loop is zero. on choosing two paths, the first is from A to B to C to D to A, and the second is from A to C to D to A. The potential difference along the axis is always zero. Potential difference for the first path is given by,

$\begin{array}{c}\mathrm{\Delta V}=({\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}})+({\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{B}})+({\mathrm{V}}_{\mathrm{D}}-{\mathrm{V}}_{\mathrm{C}})+({\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{D}})\\ =({\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}})+({\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{D}})\\ =\frac{{\mathrm{Qs}}_1}{{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2}-\frac{{\mathrm{Qs}}_1}{{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2}\\ =0\end{array}$

Potential for second path is given by,

$\begin{array}{c}\mathrm{\Delta V}=({\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{A}})+({\mathrm{V}}_{\mathrm{D}}-{\mathrm{V}}_{\mathrm{C}})+({\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{D}})\\ =({\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{A}})+({\mathrm{V}}_{\mathrm{D}}-{\mathrm{V}}_{\mathrm{C}})\\ =-{\mathrm{E}}_{\mathrm{x}}({\mathrm{x}}_{\mathrm{C}}-{\mathrm{x}}_{\mathrm{A}})-{\mathrm{E}}_{\mathrm{x}}({\mathrm{x}}_{\mathrm{D}}-{\mathrm{x}}_{\mathrm{C}})\\ =-\frac{\mathrm{Q}}{{\mathrm{A\epsilon }}_0}(-{\mathrm{s}}_1)-\frac{\mathrm{Q}}{{\mathrm{A\epsilon }}_0}\left({\mathrm{s}}_1\right)\\ =0\end{array}$

Thus, it is proved that the potential difference in any path forming a close loop is zero.