Question

In the region of space depicted in Figure 16.87 there are several stationary charged objects that are not shown in the diagram. Along a path $\mathbf{A}\mathbf{=}\mathbf{B}\mathbf{=}\mathbf{C}\mathbf{=}\mathbf{D}$ you measure the following potential differences:${\mathbf{V}}_{\mathbf{B}}\mathbf{-}{\mathbf{V}}_{\mathbf{A}}\mathbf{=}\mathbf{12}\mathbf{\text{\hspace{0.17em}V}}$ ;${\mathbf{V}}_{\mathbf{C}}\mathbf{-}{\mathbf{V}}_{\mathbf{B}}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{\text{\hspace{0.17em}V}}$ ; ${\mathit{V}}_{\mathbf{D}}\mathbf{-}{\mathit{V}}_{\mathbf{C}}\mathbf{=}\mathbf{-}\mathbf{15}\mathbf{\text{\hspace{0.17em}V}}$. What is the potential difference${\mathit{V}}_{\mathbf{A}}\mathbf{-}{\mathit{V}}_{\mathbf{D}}$ ?

Short answer

The potential difference between point A and D is $8V$.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The potential difference between point A and B is,${\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}=12\text{\hspace{0.17em}V}$
• The potential difference between point B and C is,${\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{B}}=-5\text{\hspace{0.17em}V}$
• The potential difference between point D and C is ,$V_D-V_C=-15\text{\hspace{0.17em}V}$

Significance of electric potential along a closed path

Voltage is the term used to describe the electric potential. The law of conservation states that the sum of the potential difference at all the edges in a closed path is equal to the zero.

Determination of the potential difference.

According to the law of conservation, total potential difference should be zero on the path ABCD, which can be given as follows,

$\begin{array}{c}\mathrm{\Delta V}=({\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}})+({\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{B}})+({\mathrm{V}}_{\mathrm{D}}-{\mathrm{V}}_{\mathrm{C}})+({\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{D}})\\ 0=({\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}})+({\mathrm{V}}_{\mathrm{C}}-{\mathrm{V}}_{\mathrm{B}})+({\mathrm{V}}_{\mathrm{D}}-{\mathrm{V}}_{\mathrm{C}})+({\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{D}})\end{array}$

Here,$V_B-V_A$is the potential difference between point A and B,$V_C-V_B$is the potential difference between point B and C,$V_D-V_C$ is the potential difference between point B and C, and $V_A-V_D$is the potential difference between point A and point D.

Rearrange the above expression to get the expression for the potential difference between point A and D.

$V_A-V_D=0-(V_B-V_A)-(V_C-V_B)-(V_D-V_C)$

Substitute all the values in the above expression.

$\begin{array}{c}{V}_A-V_D=-12\text{\hspace{0.17em}V}-(-5\text{\hspace{0.17em}V})-(-15\text{\hspace{0.17em}V})\\ =8\text{\hspace{0.17em}V}\end{array}$

Thus, the potential difference between point A and D is$8\text{\hspace{0.17em}V}$ .