Question

Rank the locations A to F on the basis of the electric potential at each point. Rank positive electric potentials as higher than negative electric potentials.

Rank the locations from highest to lowest potential. To rank items as equivalent, overlap them.

Answer 1

Concepts and reason

The concept required to solve this question is electric potential.

First assume that each box represents 1 unit after that calculate the electric potential of positive and negative charge by using the formula for electric potential at various points. Finally rank the potentials at various points from higher to lower potential.

Fundamentals

The electric potential surrounding a point charge is expressed as follows:

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

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the charge, r is the distance between the source charge and the point of ineptest, and V is the electric potential.

The positive charges create a positive electric potential and negative charges creates negative potentials in their vicinity. The positive charge visualizes as a elevation and negative charges visualize as a depression.

Calculate the electric potential at point A due to positive and negative charges.

The electric potential surrounding a point charge is expressed as follows:

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

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the charge, r is the distance between the source charge and the point of ineptest, and V is the electric potential.

The electric potential at point A due to positive charge is,

${V_{{\rm{Ap}}}} = \frac{{kq}}{{{r_{{\rm{Ap}}}}}}$

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the positive charge, ${r_{{\rm{Ap}}}}$ is the distance between the positive charge and the point A, and ${V_{\rm{A}}}$ is the electric potential at point A.

Substitute 2 for ${r_{{\rm{Ap}}}}$ in expression ${V_{\rm{A}}} = \frac{{kq}}{{{r_{{\rm{Ap}}}}}}$ .

${V_{{\rm{Ap}}}} = \frac{{kq}}{2}$

The electric potential at point A due to negative charge is,

${V_{{\rm{An}}}} = \frac{{kq}}{{{r_{{\rm{An}}}}}}$

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the positive charge, ${r_{{\rm{An}}}}$ is the distance between the negative charge and the point A, and ${V_{\rm{A}}}$ is the electric potential at point A.

Substitute 6 for ${r_{{\rm{Bp}}}}$ and $- q$ for q in expression ${V_{\rm{A}}} = \frac{{kq}}{{{r_{{\rm{An}}}}}}$ .

${V_{\rm{A}}}_{\rm{n}} = - \frac{{kq}}{6}$

So, the total electric potential at point A due to negative and positive charge is,

${V_{\rm{A}}} = {V_{{\rm{Ap}}}} + {V_{{\rm{An}}}}$

Substitute $\frac{{kq}}{2}$ for ${V_{{\rm{Ap}}}}$ and $- \frac{{kq}}{6}$ for ${V_{{\rm{An}}}}$ .

$\begin{array}{c}\\{V_{\rm{A}}} = \frac{{kq}}{2} - \frac{{kq}}{6}\\\\ = \frac{{kq}}{3}\\\end{array}$

Calculate the electric potential at point B due to positive and negative charges.

The electric potential surrounding a point charge is expressed as follows:

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

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the charge, r is the distance between the source charge and the point of ineptest, and V is the electric potential.

The electric potential at point B due to positive charge is,

${V_{{\rm{Bp}}}} = \frac{{kq}}{{{r_{{\rm{Bp}}}}}}$

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the positive charge, ${r_{{\rm{Bp}}}}$ is the distance between the positive charge and the point B, and ${V_{\rm{B}}}$ is the electric potential at point B.

Substitute 1 for ${r_{{\rm{Bp}}}}$ in expression ${V_{{\rm{Bp}}}} = \frac{{kq}}{{{r_{{\rm{Bp}}}}}}$ .

${V_{{\rm{Bp}}}} = \frac{{kq}}{1}$

The electric potential at point B due to negative charge is,

${V_{{\rm{Bn}}}} = \frac{{kq}}{{{r_{{\rm{Bn}}}}}}$

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the positive charge, ${r_{{\rm{Bn}}}}$ is the distance between the negative charge and the point B, and ${V_{\rm{B}}}$ is the electric potential at point B.

Substitute 5 for ${r_{{\rm{Bp}}}}$ and $- q$ for q in expression ${V_{{\rm{Bn}}}} = \frac{{kq}}{{{r_{{\rm{Bn}}}}}}$ .

${V_B}_{\rm{n}} = - \frac{{kq}}{5}$

So, the total electric potential at point B due to negative and positive charge is,

${V_{\rm{B}}} = {V_{{\rm{Bp}}}} + {V_{{\rm{Bn}}}}$

Substitute $\frac{{kq}}{1}$ for ${V_{{\rm{Bp}}}}$ and $- \frac{{kq}}{5}$ for ${V_{{\rm{Bn}}}}$ .

$\begin{array}{c}\\{V_{\rm{B}}} = \frac{{kq}}{1} - \frac{{kq}}{5}\\\\ = \frac{{4kq}}{5}\\\end{array}$

Calculate the electric potential at point E due to positive and negative charges.

The electric potential surrounding a point charge is expressed as follows:

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

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the charge, r is the distance between the source charge and the point of ineptest, and V is the electric potential.

The electric potential at point E due to positive charge is,

${V_{{\rm{Ep}}}} = \frac{{kq}}{{{r_{{\rm{Ep}}}}}}$

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the positive charge, ${r_{{\rm{Ep}}}}$ is the distance between the positive charge and the point E, and ${V_{\rm{E}}}$ is the electric potential at point E.

Substitute 5 for ${r_{{\rm{Ep}}}}$ in expression ${V_{{\rm{Ep}}}} = \frac{{kq}}{{{r_{{\rm{Ep}}}}}}$ .

${V_{{\rm{Ep}}}} = \frac{{kq}}{5}$

The electric potential at point E due to negative charge is,

${V_{{\rm{En}}}} = \frac{{kq}}{{{r_{{\rm{En}}}}}}$

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the positive charge, ${r_{{\rm{En}}}}$ is the distance between the negative charge and the point E, and ${V_{\rm{E}}}$ is the electric potential at point E.

Substitute 1 for ${r_{{\rm{Ep}}}}$ and $- q$ for q in expression ${V_{{\rm{En}}}} = \frac{{kq}}{{{r_{{\rm{En}}}}}}$ .

${V_{{\rm{En}}}} = - \frac{{kq}}{1}$

So, the total electric potential at point B due to negative and positive charge is,

${V_{\rm{E}}} = {V_{{\rm{Ep}}}} + {V_{{\rm{En}}}}$

Substitute $- \frac{{kq}}{1}$ for ${V_{{\rm{Ep}}}}$ and $\frac{{kq}}{5}$ for ${V_{{\rm{En}}}}$ .

$\begin{array}{c}\\{V_{\rm{E}}} = \frac{{kq}}{5} - \frac{{kq}}{1}\\\\ = - \frac{{4kq}}{5}\\\end{array}$

Calculate the electric potential at point f due to positive and negative charges.

The electric potential surrounding a point charge is expressed as follows:

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

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the charge, r is the distance between the source charge and the point of ineptest, and V is the electric potential.

The electric potential at point F due to positive charge is,

${V_{{\rm{Fp}}}} = \frac{{kq}}{{{r_{{\rm{Fp}}}}}}$

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the positive charge, ${r_{{\rm{Fp}}}}$ is the distance between the positive charge and the point F, and ${V_{\rm{F}}}$ is the electric potential at point F.

Substitute 6 for ${r_{{\rm{Fp}}}}$ in expression ${V_{{\rm{Fp}}}} = \frac{{kq}}{{{r_{{\rm{Fp}}}}}}$ .

${V_{{\rm{Fp}}}} = \frac{{kq}}{6}$

The electric potential at point F due to negative charge is,

${V_{{\rm{Bn}}}} = \frac{{kq}}{{{r_{{\rm{Bn}}}}}}$

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the positive charge, ${r_{{\rm{Fn}}}}$ is the distance between the negative charge and the point F, and ${V_{\rm{F}}}$ is the electric potential at point F.

Substitute 2 for ${r_{{\rm{Fp}}}}$ and $- q$ for q in expression ${V_{{\rm{Fn}}}} = \frac{{kq}}{{{r_{{\rm{Fn}}}}}}$ .

${V_{{\rm{Fn}}}} = - \frac{{kq}}{2}$

So, the total electric potential at point F due to negative and positive charge is,

${V_{\rm{F}}} = {V_{{\rm{Fp}}}} + {V_{{\rm{Fn}}}}$

Substitute $\frac{{kq}}{6}$ for ${V_{{\rm{Bp}}}}$ and $- \frac{{kq}}{2}$ for ${V_{{\rm{Bn}}}}$ .

$\begin{array}{c}\\{V_{\rm{F}}} = \frac{{kq}}{6} - \frac{{kq}}{2}\\\\ = - \frac{{kq}}{3}\\\end{array}$

Calculate the electric potential at point C due to positive and negative charges.

The electric potential surrounding a point charge is expressed as follows:

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

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the charge, r is the distance between the source charge and the point of ineptest, and V is the electric potential.

The electric potential at point F due to positive charge is,

${V_{{\rm{Cp}}}} = \frac{{kq}}{{{r_{{\rm{Cp}}}}}}$

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the positive charge, ${r_{{\rm{Cp}}}}$ is the distance between the positive charge and the point C, and ${V_{\rm{C}}}$ is the electric potential at point C.

Substitute $\sqrt {{\rm{13}}}$ for ${r_{{\rm{Cp}}}}$ in expression ${V_{{\rm{Cp}}}} = \frac{{kq}}{{{r_{{\rm{Cp}}}}}}$ .

${V_{{\rm{Cp}}}} = \frac{{kq}}{{\sqrt {{\rm{13}}} }}$

The electric potential at point F due to negative charge is,

${V_{{\rm{Cn}}}} = \frac{{kq}}{{{r_{{\rm{Cn}}}}}}$

Here, k is the constant having value of $9.0 \times {10^9}\,{\rm{N \bullet }}\frac{{{{\rm{m}}^{\rm{2}}}}}{{{{\rm{C}}^{\rm{2}}}}}$ , q is the positive charge, ${r_{{\rm{Cn}}}}$ is the distance between the negative charge and the point C, and ${V_{\rm{C}}}$ is the electric potential at point C.

Substitute $\sqrt {{\rm{13}}}$ for ${r_{{\rm{Cp}}}}$ and $- q$ for q in expression ${V_{{\rm{Cn}}}} = \frac{{kq}}{{{r_{{\rm{Cn}}}}}}$ .

${V_{{\rm{Cn}}}} = - \frac{{kq}}{{\sqrt {{\rm{13}}} }}$

So, the total electric potential at point C due to negative and positive charge is,

${V_{\rm{C}}} = {V_{{\rm{Cp}}}} + {V_{{\rm{Cn}}}}$

Substitute $\frac{{kq}}{{\sqrt {{\rm{13}}} }}$ for ${V_{{\rm{Bp}}}}$ and $- \frac{{kq}}{{\sqrt {{\rm{13}}} }}$ for ${V_{{\rm{Bn}}}}$ .

$\begin{array}{c}\\{V_{\rm{C}}} = \frac{{kq}}{{\sqrt {{\rm{13}}} }} - \frac{{kq}}{{\sqrt {{\rm{13}}} }}\\\\ = 0\\\end{array}$

Now, due to symmetry the potential at point D is equal to the potential at point C.

${V_{\rm{D}}} = 0$

Ans:

The rank from highest potential to lowest potential is $B > A > C = D > F > E$ .