Question

In the figure below, eight particles form a square, with distanced between adjacent particles. What is the electricpotential at point P at the center of the square if theelectric potential is zero at infinity?
1q/4πε₀d

***Why is the answer -4?? Please explain. Thanks****

Answer 1

Concepts and reason

The main concept used to solve the problem is electric potential due to a point charge.

Initially, Calculate the distance of the charges on the corners of the square from the center using Pythagoras theorem.

Finally, calculate the magnitude and direction of electric potential using the relation equating electric potential with charge and distance.

Fundamentals

The electric potential is a quantity numerically equal to the potential energy of a unit positive charge at a given point of the field. The formula for electric potential is,

$V = \frac{{{k_{\rm{e}}}q}}{r}$

Here, ${k_{\rm{e}}}$ is coulomb’s constant, $q$ is the charge and $r$ is the distance.

The figure given below shows the arrangement of charges around the square.

Here, $q$ is the charge and d is the distance.

From the figure given above,

In triangle XAW,

${\rm{X}}{{\rm{A}}^2} + {\rm{W}}{{\rm{A}}^2} = {\rm{X}}{{\rm{W}}^2}$

Substitute $d$ for ${\rm{XA}}$ and ${\rm{WA}}$ in the above equation.

$\begin{array}{c}\\{\rm{X}}{{\rm{W}}^2} = 2{d^2}\\\\{\rm{XW = }}\sqrt 2 d\\\end{array}$

Also, ${r_{{\rm{PA}}}} = {r_{{\rm{PB}}}} = {r_{{\rm{PC}}}} = {r_{{\rm{PD}}}} = r = {\rm{XW = }}\sqrt 2 d$

The electric potential at point P due to the charge placed at the corner of the square A is,

${V_{{\rm{PA}}}} = \frac{{{k_{\rm{e}}}{q_{\rm{A}}}}}{{{r_{{\rm{PA}}}}}}$

Here, ${k_{\rm{e}}}$ is the coulomb’s constant, ${q_{\rm{A}}}$ is the charge at point A and ${r_{{\rm{PA}}}}$ is the distance of point A from point P.

Substitute $\sqrt 2 d$ for ${r_{{\rm{PA}}}}$ and $- 4q$ for ${q_{\rm{A}}}$ in the above equation.

${V_{{\rm{PA}}}} = \frac{{ - 4{k_{\rm{e}}}q}}{{\sqrt 2 d}}$

The electric potential at point P due to the charge placed at the corner of the square B is,

${V_{{\rm{PB}}}} = \frac{{{k_{\rm{e}}}{q_{\rm{B}}}}}{{{r_{{\rm{PB}}}}}}$

Here, ${k_{\rm{e}}}$ is the coulomb’s constant, ${q_{\rm{B}}}$ is the charge at point B and ${r_{{\rm{PB}}}}$ is the distance of point B from point P.

Substitute $\sqrt 2 d$ for ${r_{{\rm{PB}}}}$ and $q$ for ${q_{\rm{B}}}$ in the above equation.

${V_{{\rm{PB}}}} = \frac{{{k_{\rm{e}}}q}}{{\sqrt 2 d}}$

The electric potential at point P due to the charge placed at the corner of the square C is,

${V_{{\rm{PC}}}} = \frac{{{k_{\rm{e}}}{q_{\rm{C}}}}}{{{r_{{\rm{PC}}}}}}$

Here, ${k_{\rm{e}}}$ is the coulomb’s constant, ${q_{\rm{C}}}$ is the charge at point C and ${r_{{\rm{PC}}}}$ is the distance of point C from point P.

Substitute $\sqrt 2 d$ for ${r_{{\rm{PC}}}}$ and $4q$ for ${q_{\rm{C}}}$ in the above equation.

${V_{{\rm{PC}}}} = \frac{{4{k_{\rm{e}}}q}}{{\sqrt 2 d}}$

The electric potential at point P due to the charge placed at the corner of the square D is,

${V_{{\rm{PD}}}} = \frac{{{k_{\rm{e}}}{q_{\rm{D}}}}}{{{r_{{\rm{PD}}}}}}$

Here, ${k_{\rm{e}}}$ is the coulomb’s constant, ${q_{\rm{D}}}$ is the charge at point D and ${r_{{\rm{PD}}}}$ is the distance of point D form point P.

Substitute $\sqrt 2 d$ for ${r_{{\rm{PD}}}}$ and $- q$ for ${q_{\rm{D}}}$ in the above equation.

${V_{{\rm{PD}}}} = \frac{{ - {k_{\rm{e}}}q}}{{\sqrt 2 d}}$

From the figure given above,

${r_{{\rm{PX}}}} = {r_{{\rm{PY}}}} = {r_{{\rm{PZ}}}} = {r_{{\rm{PW}}}} = d$

The electric potential at point P due to the charge placed at point X is,

${V_{{\rm{PX}}}} = \frac{{{k_{\rm{e}}}{q_{\rm{X}}}}}{{{r_{{\rm{PX}}}}}}$

Here, ${k_{\rm{e}}}$ is the coulomb’s constant, ${q_{\rm{X}}}$ is the charge at point X and ${r_{{\rm{PX}}}}$ is the distance of point X from point P.

Substitute $d$ for ${r_{{\rm{PX}}}}$ and $- 2q$ for ${q_{\rm{X}}}$ in the above equation.

${V_{{\rm{PX}}}} = \frac{{ - 2{k_{\rm{e}}}q}}{d}$

The electric potential at point P due to the charge placed at point Y is,

${V_{{\rm{PY}}}} = \frac{{{k_{\rm{e}}}{q_{\rm{Y}}}}}{{{r_{{\rm{PY}}}}}}$

Here, ${k_{\rm{e}}}$ is the coulomb’s constant, ${q_{\rm{Y}}}$ is the charge at point Y and ${r_{{\rm{PY}}}}$ is the distance of point Y from point P.

Substitute $d$ for ${r_{{\rm{PY}}}}$ and $- 5q$ for ${q_{\rm{Y}}}$ in the above equation.

${V_{{\rm{PY}}}} = \frac{{ - 5{k_{\rm{e}}}q}}{d}$

The electric potential at point P due to the charge placed at point Z is,

${V_{{\rm{PZ}}}} = \frac{{{k_{\rm{e}}}{q_{\rm{Z}}}}}{{{r_{{\rm{PZ}}}}}}$

Here, ${k_{\rm{e}}}$ is the coulomb’s constant, ${q_{\rm{Z}}}$ is the charge at point C and ${r_{{\rm{PZ}}}}$ is the distance of point Z from point P.

Substitute $d$ for ${r_{{\rm{PZ}}}}$ and $- 2q$ for ${q_{\rm{Z}}}$ in the above equation.

${V_{{\rm{PZ}}}} = \frac{{ - 2{k_{\rm{e}}}q}}{d}$

The electric potential at point P due to the charge placed at point W is,

${V_{{\rm{PW}}}} = \frac{{{k_{\rm{e}}}{q_{\rm{W}}}}}{{{r_{{\rm{PW}}}}}}$

Here, ${k_{\rm{e}}}$ is the coulomb’s constant, ${q_{\rm{W}}}$ is the charge at point W and ${r_{{\rm{PW}}}}$ is the distance of point W form point P.

Substitute $d$ for ${r_{{\rm{PW}}}}$ and $5q$ for ${q_{\rm{W}}}$ in the above equation.

${V_{{\rm{PW}}}} = \frac{{5{k_{\rm{e}}}q}}{d}$

The total potential at point P due to all the charges is,

${V_{{\rm{tot}}}} = {V_{{\rm{PA}}}} + {V_{{\rm{PB}}}} + {V_{{\rm{PC}}}} + {V_{{\rm{PD}}}} + {V_{{\rm{PX}}}} + {V_{{\rm{PY}}}} + {V_{{\rm{PZ}}}} + {V_{{\rm{PW}}}}$

Substitute $\frac{{ - 4{k_{\rm{e}}}q}}{{\sqrt 2 d}}$ for ${V_{{\rm{PA}}}}$ , $\frac{{{k_{\rm{e}}}q}}{{\sqrt 2 d}}$ for ${V_{{\rm{PB}}}}$ , $\frac{{4{k_{\rm{e}}}q}}{{\sqrt 2 d}}$ for ${V_{{\rm{PC}}}}$ , $\frac{{ - {k_{\rm{e}}}q}}{{\sqrt 2 d}}$ for ${V_{{\rm{PD}}}}$ , $\frac{{ - 2{k_{\rm{e}}}q}}{d}$ for ${V_{{\rm{PX}}}}$ , $\frac{{ - 5{k_{\rm{e}}}q}}{d}$ for ${V_{{\rm{PY}}}}$ , $\frac{{ - 2{k_{\rm{e}}}q}}{d}$ for ${V_{{\rm{PZ}}}}$ and $\frac{{5{k_{\rm{e}}}q}}{d}$ for ${V_{{\rm{PW}}}}$ in the above equation.

$\begin{array}{c}\\{V_{{\rm{tot}}}} = {V_{{\rm{PA}}}} + {V_{{\rm{PB}}}} + {V_{{\rm{PC}}}} + {V_{{\rm{PD}}}} + {V_{{\rm{PX}}}} + {V_{{\rm{PY}}}} + {V_{{\rm{PZ}}}} + {V_{{\rm{PW}}}}\\\\ = \frac{{ - 4{k_{\rm{e}}}q}}{{\sqrt 2 d}} + \frac{{{k_{\rm{e}}}q}}{{\sqrt 2 d}} + \frac{{4{k_{\rm{e}}}q}}{{\sqrt 2 d}} + \frac{{ - {k_{\rm{e}}}q}}{{\sqrt 2 d}} + \frac{{ - 2{k_{\rm{e}}}q}}{d} + \frac{{ - 5{k_{\rm{e}}}q}}{d} + \frac{{ - 2{k_{\rm{e}}}q}}{d} + \frac{{5{k_{\rm{e}}}q}}{d}\\\end{array}$

${V_{{\rm{tot}}}} = \frac{{ - 4{k_{\rm{e}}}q}}{d}$ …… (1)

Also, the coulomb’s constant is given by,

${k_{\rm{e}}} = \frac{1}{{4\pi {\varepsilon _0}}}$

Substitute $\frac{1}{{4\pi {\varepsilon _0}}}$ for ${k_{\rm{e}}}$ in equation (1).

${V_{{\rm{tot}}}} = - 4\left( {\frac{q}{{4\pi {\varepsilon _0}d}}} \right)$

Ans:

The electric potential at point P is $- 4\left( {\frac{q}{{4\pi {\varepsilon _0}d}}} \right)$ .