Q.[Case study] Consider the charges q1 and q2 initially at infinity and determine the work done by an external agency to bring the charges to the given locations. Suppose, first the charge q1 is brought from infinity to the point r1⃗. There is no external field against which work needs to be done, so work done in bringing q1 from infinity to r1⃗ is zero. From the definition of potential, work done in bringing charge q2 from infinity to the point r2⃗ is q2 times the potential at r2⃗ due to q1. (Figure: two point charges q1 and q2 joined by a line of length r12.)

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:

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

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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:

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

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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:

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

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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:

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

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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 $k = 9\times10^{9}\,N\,m^2/C^2$.

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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.

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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 $\dfrac{1}{r^3}$:

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

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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

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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.

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