Q.Determine the electrostatic potential energy of a system with no external field, consisting of two charges $7\mu C$ and $-2\mu C$ placed at $(-9\text{ cm}, 0, 0)$ and $(+9\text{ cm}, 0, 0)$ respectively.

The potential energy of a system of point charges is the sum of the potential energies of every distinct pair. For the given charges $2q$, $-2q$, and $q$ at the vertices of an equilateral triangle of side $l$, the total potential energy comes out to $-\dfrac{q^2}{\pi\varepsilon_0 l}$, which corresponds to option (D).

Two positive alpha particles repel; pushing them together raises the potential energy.

An alpha particle carries charge $+2e$. The electrostatic potential energy of two point charges is:

$U=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r}$

…

The work done to bring $q_2$ from infinity is just $q_2$ times the potential already set up by $q_1$ at that point — no squared or cubed distance term.

Since $q_1$ is already placed at $\vec r_1$, it sets up an electrostatic potential everywhere in space. At the point $\vec r_2$ (distance $r_{12}$ away from $q_1$), this potential is:

$V_1(\vec r_2) = \frac{1}{4\pi\varepsilon_0}\frac{q_1}{r_{12}}$

The work done in bringing charge $q_2$ from infinity to $\vec r_2$ against this potential (by definition of electric potential, $V = W/q$) is:

$W_2 = q_2 \, V_1(\vec r_2) = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{12}}$

…

The work to bring the third charge is $q_3$ times the potential already created at that point by BOTH earlier charges combined.

By the time $q_3$ is brought in, both $q_1$ (at $\vec r_1$) and $q_2$ (at $\vec r_2$) are already in place, having together set up a potential at the location $\vec r_3$ (where $q_3$ will go):

$V_{1,2}(\vec r_3) = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_{13}} + \frac{q_2}{r_{23}}\right)$

The work done in bringing $q_3$ from infinity to this point, against this combined potential, is:

$W_3 = q_3\, V_{1,2}(\vec r_3) = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1q_3}{r_{13}} + \frac{q_2q_3}{r_{23}}\right)$

…

The total potential energy of the three-charge system is the sum of all THREE distinct pairwise interaction terms, each with a plain '+' sign.

Assembling the three charges one at a time (as in parts i–iii): $q_1$ costs no work ($W_1=0$), $q_2$ costs $W_2 = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$, and $q_3$ costs $W_3 = \dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1q_3}{r_{13}} + \dfrac{q_2q_3}{r_{23}}\right)$.

The total work done to assemble the configuration — which equals the total electrostatic potential energy of the system, since this work is stored as PE — is:

$U = W_1 + W_2 + W_3 = \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1q_2}{r_{12}} + \frac{q_1q_3}{r_{13}} + \frac{q_2q_3}{r_{23}}\right)$

…

The mutual potential energy of two point charges is U = kq₁q₂/r; plugging in gives 9×10⁻³ J.

The electric potential energy of a system of two point charges is

$U = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r} = \frac{k q_1 q_2}{r}$

Here both charges are equal and positive: $q_1 = q_2 = 1\,\mu C = 1\times10^{-6}\,C$, the separation is $r = 1\,m$, and $k = 9\times10^{9}\,N\,m^2/C^2$.

…

Potential energy of two charges is U = kq₁q₂/r; for opposite signs the product q₁q₂ is negative, so U is always negative.

The mutual potential energy of two point charges is

$U = \frac{k q_1 q_2}{r}$

For two charges of opposite sign, one is positive and one is negative, so the product $q_1 q_2 < 0$. Since k and r are positive, U is negative for every finite separation.

…

Bringing two like (positive) charges closer increases their electrostatic potential energy.

The electric potential energy of a pair of point charges is $U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r}$. Both protons carry positive charge, so as one proton is brought closer to the other, the separation $r$ decreases, and since $U\propto 1/r$ for like charges, $U$ increases. Physically, work has to be done against the mutually repuls …

Dipole–dipole interaction energy ∝ 1/r³, not r², r³ or r⁴ → 'none of these'.

A single dipole's field falls off as 1/r³. The potential energy of a second dipole placed in that field is therefore proportional to $\dfrac{1}{r^3}$:

$U \propto \frac{p_1 p_2}{r^3}$

…

The sign of the effect depends on whether the charges are like or unlike, so U may increase or decrease.

The electrostatic potential energy of two charges is

$U = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r}.$

• Like charges ($q_1q_2>0$): $U$ is positive and $\propto 1/r$, so increasing $r$ decreases $U$. …

Both charges are negative, so their product is positive, giving a positive (repulsive) potential energy — plug straight into the two-point-charge PE formula.

Formula

$U = \frac{kq_1q_2}{r}$

Substitute

…

The potential energy of two point charges separated by a distance r is U = kq1q2/r; substituting the given values gives 9x10^-3 J.

The electrostatic potential energy of a system of two point charges q1 and q2 held a distance r apart in vacuum (the work needed to assemble them, starting from infinite separation) is

U = (1/4piepsilon0) * q1q2/r = kq1*q2/r, where k = 9 x 10^9 N m^2/C^2.

…