Question

Rank the electric potentials at the four points shown in the figure below from largest to smallest. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.)

Answer 1

Concepts and reason

The concept used to solve this question is electric potential.

First, determine the distance between corresponding charges to the corresponding points by drawing the diagram and then determine the electric potential at point A and B by using the electric potential formula.

Later, determine the electric potential at the point $C$ and $D$ by using the diagram and the electric potential formula.

Fundamentals

Electric potential:

The electric potential is defined as the work required to move a unit charge from a reference point to a specified point.

The electric potential is defined as,

$V = \frac{{kq}}{r}$

Here, $V$ is the electric potential, $k$ is the Coulomb constant, $q$ is the charge and $r$ is the distance between the charged particle to the required point of electric potential.

The total electric potential at any point is calculated as the scalar sum of all potentials due to different charges.

Therefore, the electric potential at point A and B can be expressed as,

$V = \frac{{k{q_1}}}{{{r_1}}} + \frac{{k{q_2}}}{{{r_2}}}$ …… (1)

Here, ${q_1}$ and ${q_2}$ are two charges which have the value of $Q\,\,{\rm{and}}\,\,2Q$ respectively, ${r_1}$ is the distance of the corresponding point from charge $Q$ and ${r_2}$ is the distance of the corresponding point from charge $2Q$ .

The electric potential at point A is ${V_A}$ .

Substitute $d$ for ${r_1}$ , $\sqrt 2 d$ for ${r_2}$ , $Q$ for ${q_1}$ and $2Q$ for ${q_2}$ in equation (1).

$\begin{array}{c}\\{V_{\rm{A}}} = \frac{{k\left( Q \right)}}{d} + \frac{{k\left( {2Q} \right)}}{{\sqrt 2 d}}\\\\ = \frac{{kQ}}{d}\left( {1 + \frac{2}{{\sqrt 2 }}} \right)\\\\ = \left( {2.41} \right)\frac{{kQ}}{d}\\\end{array}$

The electric potential at point B is ${V_B}$ .

Substitute $\sqrt 2 d$ for ${r_1}$ , $d$ for ${r_2}$ , $Q$ for ${q_1}$ and $2Q$ for ${q_2}$ in equation (1).

$\begin{array}{c}\\{V_{\rm{B}}} = \frac{{k\left( Q \right)}}{{\sqrt 2 d}} + \frac{{k\left( {2Q} \right)}}{d}\\\\ = \frac{{kQ}}{d}\left( {\frac{1}{{\sqrt 2 }} + 2} \right)\\\\ = \left( {2.71} \right)\frac{{kQ}}{d}\\\end{array}$

The electric potential at point C is ${V_C}$

Substitute $\left( {\frac{{\sqrt 2 d}}{2}} \right)$ for ${r_1}$ , $\left( {\frac{{\sqrt 2 d}}{2}} \right)$ for ${r_2}$ , $Q$ for ${q_1}$ and $2Q$ for ${q_2}$ in equation (1).

$\begin{array}{c}\\{V_{\rm{C}}} = \frac{{k\left( Q \right)}}{{\left( {\frac{{\sqrt 2 d}}{2}} \right)}} + \frac{{k\left( {2Q} \right)}}{{\left( {\frac{{\sqrt 2 d}}{2}} \right)}}\\\\ = \frac{{kQ}}{d}\left( {\frac{2}{{\sqrt 2 }} + \frac{4}{{\sqrt 2 }}} \right)\\\\ = \left( {4.24} \right)\frac{{kQ}}{d}\\\end{array}$

The electric potential at point D is ${V_D}$

Substitute $\left( {\frac{d}{2}} \right)$ for ${r_1}$ , $\left( {\frac{d}{2}} \right)$ for ${r_2}$ , $Q$ for ${q_1}$ and $2Q$ for ${q_2}$ in equation (1).

$\begin{array}{c}\\{V_{\rm{D}}} = \frac{{k\left( Q \right)}}{{\left( {\frac{d}{2}} \right)}} + \frac{{k\left( {2Q} \right)}}{{\left( {\frac{d}{2}} \right)}}\\\\ = \frac{{kQ}}{d}\left( {2 + 4} \right)\\\\ = \left( 6 \right)\frac{{kQ}}{d}\\\end{array}$

From the above explanation, it clearly shows that the electric potential at point D is the highest and at point A is the lowest.

Ans:

The order of the electric potential at the points from largest to smallest is $D > C > B > A$ .