(a)(i) Draw equipotential surfaces for an electric dipole. (ii) Two point charges $q_1$ and $q_2$ are located at $\vec{r}_1$ and $\vec{r}_2$ respectively in an external electric field $\vec{E}$. Obtain an expression for the potential energy of the system. (iii) The dipole moment of a molecule is $10^{-30}\ \text{C m}$. It is placed in an electric field $\vec{E}$ of $10^{5}\ \text{V/m}$ such that its axis is along the electric field. The direction of $\vec{E}$ is suddenly changed by $60^\circ$ at an instant. Find the change in the potential energy of the dipole, at that instant.

(ii) $U = q_1 V(\vec{r}_1) + q_2 V(\vec{r}_2) + \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{12}}$. (iii) $\Delta U = 5 \times 10^{-26}\ \text{J}$.

CBSE Class 12 · 2024 · ElectrostaticsLong Answer

(a)(i) Draw equipotential surfaces for an electric dipole. (ii) Two point charges

q_{1}

and

q_{2}

are located at

r_{1}

and

r_{2}

respectively in an external electric field

E

. Obtain an expression for the potential energy of the system. (iii) The dipole moment of a molecule is

1 0^{- 30} C m

. It is placed in an electric field

E

1 0^{5} V/m

such that its axis is along the electric field. The direction of

E

is suddenly changed by

6 0^{\circ}

at an instant. Find the change in the potential energy of the dipole, at that instant.

(i) The dipole equipotentials are symmetric about the axis (positive near

+ q

, negative near

- q

), with the perpendicular-bisector plane at zero potential.
(ii) Potential energy of two charges in an external field:

U = q_{1} V (r_{1}) + q_{2} V (r_{2}) + \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r _{12}}

(iii)

U = - pE cos θ

.
Initially

θ = 0^{\circ}

U_{i} = - pE

.
Finally

θ = 6 0^{\circ}

U_{f} = - pE cos 6 0^{\circ} = - \frac{1}{2} pE

Δ U = U_{f} - U_{i} = \frac{1}{2} pE = \frac{1}{2} \times 1 0^{- 30} \times 1 0^{5}

Δ U = 5 \times 1 0^{- 26} J

CBSE Class 12 Board · 2024 · Subjective