Electric Potential Energy & Potential – Concept Booster | Class 12 Physics CBSE

By definition, the potential everywhere on an equipotential surface is the same. Therefore, $V_A = V_B$, which means the potential difference $\Delta V = 0$. Since Work Done $W = q_0 \times \Delta V$, the work done is zero.

The total potential energy of the system is the sum of energies of all three pairs.
$U_{system} = \frac{1}{4\pi\varepsilon_0} \left[ \frac{(+q)(-q)}{a} + \frac{(-q)(+q)}{a} + \frac{(+q)(+q)}{a} \right]$
$U_{system} = \frac{1}{4\pi\varepsilon_0} \left[ \frac{-q^2}{a} – \frac{q^2}{a} + \frac{q^2}{a} \right] = -\frac{1}{4\pi\varepsilon_0} \frac{q^2}{a}$.
The work done to dissociate the system is the energy required to bring them to infinity, which is equal and opposite to the binding energy: $W = -U_{system} = +\frac{1}{4\pi\varepsilon_0} \frac{q^2}{a}$.

When a dipole is placed in a uniform electric field, it experiences a torque given by $\tau = pE \sin\theta$.
The work done in rotating the dipole against this torque by a small angle $d\theta$ is $dW = \tau \, d\theta = pE \sin\theta \, d\theta$.
Total work done to rotate it from $\theta_1 = 90^\circ$ (reference point of zero potential energy) to an angle $\theta$ is:
$W = \int_{90^\circ}^{\theta} pE \sin\theta \, d\theta = pE [-\cos\theta]_{90^\circ}^{\theta} = -pE (\cos\theta – \cos90^\circ)$.
Since $\cos90^\circ = 0$, the work done is stored as potential energy: $U = -pE \cos\theta = -\vec{p} \cdot \vec{E}$.