Potential, Potential Difference and Potential Energy

Potential at a point is the work done by an external agent in bringing a unit positive test charge from infinity to that point, without acceleration.

Derivation. Let $Q$ be a point charge at $O$ in free space, and let $P$ be a point at distance $r$ from $O$. Bring a unit positive charge from infinity to $P$ along the line joining them.

At distance $x$ from $O$, the field is

$E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{x^2}$

directed radially outward for positive $Q$. The external force on the unit charge is $-E$ (opposite to the field), and the displacement is $dx$ (with $x$ decreasing from $\infty$ to $r$). The work done by the external agent is

$$V=\int_{\infty}^{r}(-E)\,dx=-\frac{Q}{4\pi\varepsilon_0}\int_{\infty}^{r}\frac{dx}{x^2} =-\frac{Q}{4\pi\varepsilon_0}\left[-\frac1x\right]_{\infty}^{r}$$

$\boxed{V=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r}}$

Potential difference. The potentials at distances $r_1$ and $r_2$ are $V_1=\dfrac{Q}{4\pi\varepsilon_0 r_1}$ and $V_2=\dfrac{Q}{4\pi\varepsilon_0 r_2}$. So

$V_1-V_2=\frac{Q}{4\pi\varepsilon_0}\left(\frac1{r_1}-\frac1{r_2}\right)$

For positive $Q$ and $r_2>r_1$, $V_1>V_2$: the potential falls with distance from the charge.

Vector or scalar?

• Electric field at a point is the force per unit positive test charge, $\vec E=\vec F/q_0$. Force is a vector, so the electric field is a vector quantity (it has magnitude and direction).
• Electric potential at a point is the work done per unit positive charge in bringing it from infinity to that point, $V=W/q_0$. Work is a scalar, so electric potential is a scalar quantity (only magnitude, with a sign). Potentials due to several charges add algebraically.

So only the electric field is a vector; the potential is not.

Potential due to a point charge

Let a charge $+q$ be at O. We bring a unit positive test charge from infinity to a point P at distance $r$. At distance $x$ the field is $E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{x^2}$, directed away from $q$. The work done against it:

$V=-\int_{\infty}^{r}\vec E\cdot d\vec x=-\int_\infty^{r}\frac{1}{4\pi\varepsilon_0}\frac{q}{x^2}\,dx=\frac{q}{4\pi\varepsilon_0}\left[\frac1x\right]_{\infty}^{r}$

$\boxed{V=\frac{1}{4\pi\varepsilon_0}\frac qr}$

It is positive for $+q$, negative for $-q$, and falls as $1/r$. Unit: volt (J C$^{-1}$).

Let $M$ be the midpoint of $AB$. Charges $A$ and $B$ are each $5\ \text{cm}=0.05\ \text{m}$ from $M$. Charge $C$ is at the height of the equilateral triangle:

$CM=\frac{\sqrt3}{2}\times0.10=0.0866\ \text{m}$

Potential is a scalar, so the contributions add:

$V=kq\left(\frac{1}{0.05}+\frac{1}{0.05}+\frac{1}{0.0866}\right)$

With $kq=9\times10^9\times1.8\times10^{-6}=1.62\times10^{4}\ \text{V m}$,

$V=1.62\times10^{4}\,(20+20+11.55)=1.62\times10^{4}\times51.55\approx 8.35\times10^{5}\ \text{V}$

Yes. Potential is a scalar, so the contributions of different charges can cancel, whereas the electric field is a vector sum that may not cancel.

Example: take an electric dipole of charges $+q$ and $-q$ separated by $2a$. At the midpoint of the line joining them (or anywhere on the perpendicular bisector) the potentials of the two charges are equal and opposite, so $V=0$. At the midpoint, however, both fields point from $+q$ towards $-q$ and add:

$E=2\times\frac{1}{4\pi\varepsilon_0}\frac{q}{a^2}\neq0$

Let $P$ be the midpoint of $BC$. Then $PB=PC=0.5\ \text{m}$ and, from the altitude of the equilateral triangle,

$AP=\sqrt{1^2-0.5^2}=0.866\ \text{m}.$

Potential is a scalar, so the potentials add:

$V=\frac{1}{4\pi\varepsilon_0}\left[\frac{q_A}{AP}+\frac{q_B}{PB}+\frac{q_C}{PC}\right]$

With $\dfrac{1}{4\pi\varepsilon_0}=\dfrac{1}{4\pi\times8.85\times10^{-12}}=8.99\times10^{9}\ \text{N m}^2\text{C}^{-2}$:

$V=8.99\times10^{9}\left[\frac{1\times10^{-6}}{0.866}+\frac{2\times10^{-6}}{0.5}+\frac{3\times10^{-6}}{0.5}\right]$

$V=8.99\times10^{9}\times(1.155+4+6)\times10^{-6}=8.99\times10^{9}\times11.155\times10^{-6}\approx1.0\times10^{5}\ \text{V}$

(d) 0 V/m

$E=-\frac{dV}{dx}=-(10x-10)$

At $x=1\ \text{m}$: $E=-(10-10)=0\ \text{V/m}$.

(b) ( \text{J/C} )

Electric potential (V = W/q), so its SI unit is joule per coulomb (J/C), which is given the name volt (V). ( \text{N/C} ) is the unit of electric field intensity.

Electric potential: The electric potential at a point in an electric field is defined as the work done per unit positive test charge in bringing it from infinity to that point against the electric field.
If (W) is the work done in bringing a charge (q_0) from infinity to a point, then electric potential is

[ V=\frac{W}{q_0} ]

It is a scalar quantity. Its SI unit is volt (V), where (1\text{ V}=1\text{ J C}^{-1}).

Electric field intensity: The electric field intensity at a point is defined as the electric force experienced per unit positive test charge placed at that point.
If a test charge (q_0) experiences force (\vec F), then

[ \vec E=\frac{\vec F}{q_0} ]

It is a vector quantity. Its SI unit is (\text{N C}^{-1}) or (\text{V m}^{-1}).

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Relation between electric field and electric potential

Consider two points (A) and (B) very close to each other in an electric field, separated by a small distance (dr) along the radial direction (\hat r). Let the potentials at (A) and (B) be (V) and (V+dV).

The work done by the electric field in moving a unit positive charge from (A) to (B) is

[ \vec E\cdot d\vec r ]

But the potential difference is

[ V_B-V_A = -\vec E\cdot d\vec r ]

So,

[ dV=-\vec E\cdot d\vec r ]

Since (d\vec r=dr,\hat r),

[ dV=-E_r,dr ]

where (E_r) is the component of electric field along (\hat r). Hence,

[ E_r=-\frac{dV}{dr} ]

Therefore, in vector form,

[ \boxed{\vec E=-\frac{dV}{dr}\hat r} ]

Thus, electric field intensity is the negative gradient of electric potential. The negative sign shows that electric field is directed from higher potential to lower potential.

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Calculation

Given,

[ V=3x^2+2y ]

The electric field is

[ \vec E=-\nabla V ]

[ \vec E=-\left(\frac{\partial V}{\partial x}\hat i+\frac{\partial V}{\partial y}\hat j+\frac{\partial V}{\partial z}\hat k\right) ]

Now,

[ \frac{\partial V}{\partial x}=6x,\qquad \frac{\partial V}{\partial y}=2,\qquad \frac{\partial V}{\partial z}=0 ]

So,

[ \vec E=-6x,\hat i-2,\hat j ]

At the point ((1,2)),

[ \vec E=-6(1)\hat i-2\hat j ]

[ \boxed{\vec E=-6\hat i-2\hat j\ \text{N C}^{-1}} ]

Its magnitude is

[ E=\sqrt{(-6)^2+(-2)^2} ]

[ E=\sqrt{36+4}=\sqrt{40}=2\sqrt{10} ]

[ \boxed{E\approx 6.32\ \text{N C}^{-1}} ]

So, the electric field at ((1,2)) is ((-6\hat i-2\hat j)\ \text{N C}^{-1}), having magnitude (6.32\ \text{N C}^{-1}).

Electric potential energy of a system of two point charges

The electric potential energy of a system of two point charges is the work done by an external agent in assembling the system by bringing the charges from infinity to their final positions quasi-statically, without acceleration. The potential energy at infinity is taken as zero. It is also the energy stored in the mutual electric field of the charges.

Derivation

Let two point charges (q_1) and (q_2) be placed in vacuum at points (A) and (B), separated by a distance (r).

1. First bring (q_1) from infinity to (A). Since there is no electric field at (A) initially, the work done is
   [ W_1=0 ]

2. Now bring (q_2) from infinity to (B) in the electric field of (q_1). At any instant, let the separation between the charges be (x). The electric force on (q_2) due to (q_1) is
   [ F_e=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{x^2} ] For quasi-static motion, the external force applied must be equal and opposite to the electric force: [ F_{\text{ext}}=-F_e=-\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{x^2} ]

3. Work done by the external agent in bringing (q_2) from infinity to distance (r) is
   [ W_2=\int_{\infty}^{r} F_{\text{ext}},dx ] [ W_2=\int_{\infty}^{r} -\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{x^2},dx ] [ W_2=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r} ]

Therefore, the electric potential energy of the system of two point charges is
[ \boxed{U=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r}} ] or
[ \boxed{U=\frac{kq_1q_2}{r}} ] where (k=\dfrac{1}{4\pi\varepsilon_0}=9\times 10^9\ \text{N m}^2\text{C}^{-2}).

If (q_1q_2>0), (U) is positive; if (q_1q_2<0), (U) is negative. The potential energy is mutual, so the same result is obtained if (q_1) is brought into the field of (q_2).

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Work done in assembling three charges

Given: [ q_1=2,\mu\text{C}=2\times 10^{-6},\text{C} ] [ q_2=-3,\mu\text{C}=-3\times 10^{-6},\text{C} ] [ q_3=4,\mu\text{C}=4\times 10^{-6},\text{C} ] [ r=10,\text{cm}=0.10,\text{m} ] [ k=9\times 10^9\ \text{N m}^2\text{C}^{-2} ]

The charges are placed at the vertices of an equilateral triangle, so the distance between every pair is (r).

Total potential energy of the system is
[ U=U_{12}+U_{23}+U_{31} ] [ U=\frac{k}{r}\left(q_1q_2+q_2q_3+q_3q_1\right) ]

Now, [ q_1q_2=(2\times 10^{-6})(-3\times 10^{-6})=-6\times 10^{-12},\text{C}^2 ] [ q_2q_3=(-3\times 10^{-6})(4\times 10^{-6})=-12\times 10^{-12},\text{C}^2 ] [ q_3q_1=(4\times 10^{-6})(2\times 10^{-6})=8\times 10^{-12},\text{C}^2 ]

So, [ q_1q_2+q_2q_3+q_3q_1=(-6-12+8)\times 10^{-12} ] [ =-10\times 10^{-12},\text{C}^2 ]

Thus, [ U=\frac{(9\times 10^9)(-10\times 10^{-12})}{0.10} ] [ U=\frac{-0.09}{0.10} ] [ \boxed{U=-0.90\ \text{J}} ]

So, the work done by the external agent in assembling the three charges is
[ \boxed{-0.90\ \text{J}} ]

The negative sign shows that net work is done by the electric field, i.e. the system releases (0.90\ \text{J}) of energy during assembly. Equivalently, the work done by the electric field is
[ \boxed{+0.90\ \text{J}} ]

Potential gradient: the rate of change of electric potential with distance in a given direction, $\dfrac{dV}{dx}$. Its unit is V m$^{-1}$.

Relation with electric field: The work done by the field in moving a small positive test charge $q_0$ through a small distance $dx$ is $dW=q_0E\,dx$. This equals the decrease in potential energy, $-q_0\,dV$.

$q_0E\,dx=-q_0\,dV\ \Rightarrow\ E=-\frac{dV}{dx}$

So the electric field intensity equals the negative of the potential gradient. The negative sign shows that the field points in the direction in which the potential decreases.