CBSE 2025, Set 2

Question

(b) (i) Obtain an expression for the electric field due to a dipole of dipole moment $\vec{p}$ at a point on its equatorial plane and specify its direction. Hence, find the value of electric field: (I) at the centre of the dipole $(r=0)$, and (II) at a point $r >> a$, where $2a$ is the length of the dipole.
(ii) An electric field $\vec{E} = (10x + 5)\hat{i}	ext{ N/C}$ exists in a region in which a cube of side $L$ is kept as shown in the figure. Here $x$ and $L$ are in metres. Calculate the net flux through the cube.

![Cube in Electric Field](https://www.edvaya.com/images/cbse/physics/2025/set2/q32b-cube-electric-field.webp)

Edvaya · Accepted Answer

The correct answer is Subjective. **(i) Electric Field on Equatorial Plane of Dipole:**

![Dipole Equatorial Derivation](https://www.edvaya.com/images/cbse/physics/2025/set2/q32b-dipole-equatorial-derivation-answer.webp)

Consider a dipole with charges $+q$ and $-q$ separated by $2a$. Point P is on the equatorial line at a distance $r$ from the centre.
The distance of P from either charge is $\sqrt{r^2 + a^2}$.
The magnitudes of the electric field at P due to $+q$ and $-q$ are equal:
$$E_{+q} = E_{-q} = \frac{1}{4\pi\epsilon_0} \frac{q}{(r^2 + a^2)}$$

The vertical components ($E_{+q}\sin	heta$ and $E_{-q}\sin	heta$) are equal and opposite, so they cancel out.
The horizontal components add up along the direction anti-parallel to the dipole moment $\vec{p}$ (which points from $-q$ to $+q$).
Total electric field magnitude $E = 2 E_{+q} \cos	heta$.
From geometry, $\cos	heta = \frac{a}{\sqrt{r^2 + a^2}}$.
$$E = 2 \left( \frac{1}{4\pi\epsilon_0} \frac{q}{r^2 + a^2} ight) \left( \frac{a}{(r^2 + a^2)^{1/2}} ight) = \frac{1}{4\pi\epsilon_0} \frac{q(2a)}{(r^2 + a^2)^{3/2}}$$
Since $p = q(2a)$, in vector form:
$$\boxed{\vec{E} = -\frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{(r^2 + a^2)^{3/2}}}$$
The direction is **opposite** to the dipole moment vector $\vec{p}$.

**(I) At the centre of the dipole ($r = 0$):**
$$\vec{E} = -\frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{(0 + a^2)^{3/2}} = \boxed{-\frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{a^3}}$$

**(II) At a far point ($r >> a$):**
We can neglect $a^2$ in the denominator.
$$\vec{E} = -\frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{(r^2)^{3/2}} = \boxed{-\frac{1}{4\pi\epsilon_0} \frac{\vec{p}}{r^3}}$$

**(ii) Net Flux through the Cube:**
The electric field $\vec{E} = (10x + 5)\hat{i}$ is along the x-axis only.
Flux through the top, bottom, front, and back faces is zero because the area vector is perpendicular to $\vec{E}$ ($	heta = 90^\circ, \cos 90^\circ = 0$).

* **Left face** ($x = 0$): Area vector is $-L^2 \hat{i}$. 
 Electric field at $x = 0$ is $E_L = (10(0) + 5)\hat{i} = 5\hat{i}$.
 Flux $\Phi_L = \vec{E_L} \cdot \vec{A_L} = (5\hat{i}) \cdot (-L^2 \hat{i}) = -5L^2$.

* **Right face** ($x = L$): Area vector is $+L^2 \hat{i}$.
 Electric field at $x = L$ is $E_R = (10L + 5)\hat{i}$.
 Flux $\Phi_R = \vec{E_R} \cdot \vec{A_R} = (10L + 5)\hat{i} \cdot (L^2 \hat{i}) = (10L + 5)L^2$.

* **Net Flux** $\Phi_{net} = \Phi_L + \Phi_R$:
 $$\Phi_{net} = -5L^2 + (10L + 5)L^2 = -5L^2 + 10L^3 + 5L^2$$
 $$\boxed{\Phi_{net} = 10L^3 	ext{ Nm}^2	ext{/C}}$$ 
 *(assuming standard units, though the prompt states L is in meters)*

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Physics > Electrostatics And Current Electricity > Electric Charges And Fields

Keep Practicing

Current Question

Community Stats

Track Your Progress

Edvaya Target

We Value Your Privacy