CBSE 2025 · Region 6 · Set 2 · Q32 · 5 marks
(i)The electric field in a region is given by $\displaystyle \overrightarrow{\mathrm{E}}=40 \mathrm{x} \hat{\mathrm{i}} \mathrm{N} / \mathrm{C}$. Find the amount of work done in taking a unit positive charge from a point $\displaystyle (0,3 m)$ to the point $\displaystyle (5 m, 0)$.(ii)A charge Q is distributed over two concentric hollow spheres of radii $\displaystyle \mathrm{r}$ and $\displaystyle R(>\mathrm{r})$ such that their surface charge densities are equal. Find :(I)the electric field, and(II)the potential at their common centre.OR 回回
(b) (i) Obtain an expression for the electric field $\displaystyle \vec{\mathrm{E}}$ due to a dipole of dipole moment $\displaystyle \vec{p}$ 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 ( $\displaystyle \mathrm{r}=0$ ), and(II)at a point $\displaystyle \mathrm{r} \gg \mathrm{a}$, where $\displaystyle 2$ a is the length of the dipole.(ii)An electric field $\displaystyle \overrightarrow{\mathrm{E}}=(10 \mathrm{x}+5) \hat{\mathrm{i}} \mathrm{N} / \mathrm{C}$ 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. 
(i)
The electric field in a region is given by $\displaystyle \overrightarrow{\mathrm{E}}=40 \mathrm{x} \hat{\mathrm{i}} \mathrm{N} / \mathrm{C}$. Find the amount of work done in taking a unit positive charge from a point $\displaystyle (0,3 m)$ to the point $\displaystyle (5 m, 0)$.
(ii)
A charge Q is distributed over two concentric hollow spheres of radii $\displaystyle \mathrm{r}$ and $\displaystyle R(>\mathrm{r})$ such that their surface charge densities are equal. Find :
(I)
the electric field, and
(II)
the potential at their common centre.
OR 回回
(b) (i) Obtain an expression for the electric field $\displaystyle \vec{\mathrm{E}}$ due to a dipole of dipole moment $\displaystyle \vec{p}$ 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 ( $\displaystyle \mathrm{r}=0$ ), and
(II)
at a point $\displaystyle \mathrm{r} \gg \mathrm{a}$, where $\displaystyle 2$ a is the length of the dipole.
(ii)
An electric field $\displaystyle \overrightarrow{\mathrm{E}}=(10 \mathrm{x}+5) \hat{\mathrm{i}} \mathrm{N} / \mathrm{C}$ 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.
Marking-scheme solution
(i)
$\displaystyle V = -\int \vec{E}\cdot \vec{dr}$
$\displaystyle = -\int 40\,x\,dx$
$\displaystyle = -20x^2$
Potential at A ($\displaystyle 0$, 3m), $\displaystyle V_A = 0$
Potential at B (5m, $\displaystyle 0$), $\displaystyle V_B = -500$ V
Work done in taking a unit positive charge from a point ($\displaystyle 0$, 3m) to the point (5m, $\displaystyle 0$)
$\displaystyle W = q(V_B - V_A)$
$\displaystyle = 1(-500 - 0)$
$\displaystyle W = -500\ \text{J}$
(ii)
(I)
Electric field at the common centre will be zero as the charge enclosed by the inner sphere is zero.
Alternatively: $\displaystyle q_{en} = 0$
$\displaystyle \phi_E = 0$
$\displaystyle \oint \vec{E}.\vec{ds} = 0$
$\displaystyle E = 0$
(II)
$\displaystyle \because$ Surface charge densities are equal
\[\frac{q}{4\pi r^2} = \frac{Q-q}{4\pi R^2}\]
\[q = \frac{Qr^2}{R^2 + r^2}\]
Potential at common centre
\[V = \frac{kq}{r} + \frac{k(Q-q)}{R}\]
\[V = \frac{k}{r}\frac{Qr^2}{(R^2+r^2)} + \frac{k}{R}\left[Q - \frac{Qr^2}{(R^2+r^2)}\right]\]
\[V = \frac{kQ(R+r)}{R^2+r^2}\]
(i)
The magnitudes of the electric field due to two charges +q and -q are
\[E_{+q} = \frac{q}{4\pi\varepsilon_0}\frac{1}{\left(r^2+a^2\right)}\]
\[E_{-q} = \frac{q}{4\pi\varepsilon_0}\frac{1}{\left(r^2+a^2\right)}\]
The total electric field
\[\vec{E} = -\left(E_{+q} + E_{-q}\right)\cos\theta\ \hat{p}\]
\[\vec{E} = -\frac{\vec{p}}{4\pi\varepsilon_0\left(r^2+a^2\right)^{3/2}}\]
Direction of electric field is opposite to dipole moment ($\displaystyle \vec{p}$)
(I)
At centre of dipole, r = $\displaystyle 0$
\[\vec{E} = -\frac{-\vec{p}}{4\pi\varepsilon_0 a^3}\]
(II)
At a point r>>a
\[\vec{E} = -\frac{-\vec{p}}{4\pi\varepsilon_0 r^3}\]
(ii)
$\displaystyle \vec{E} = (10x+5)\hat{i}$ N/C
\[\phi_L = \int \vec{E}.\vec{ds}\]
\[= -E_L(L^2)\]
\[= -5L^2\]
\[\phi_R = E_R(L^2)\]
\[= (10L+5)L^2\]
\[\phi_{net} = \phi_L + \phi_R\]
\[= -5L^2 + (10L+5)L^2\]
\[= 10L^3\ \text{Nm}^2/\text{C}\]
Electric Charges and FieldsGauss’s LawApplylong_answerhard
More from Electric Charges and Fields
- (i) Two point charges +q and -q are held at (a, 0) and ( -a, 0 ) in x-y plane. Obtain an expression for the…2025 · asked 3×
- An electric dipole consists of two point charges q and -q separated by a distance 2a. Derive an expression…2026 · asked 3×
- (i) Define electric flux and write its SI unit. (ii) Use Gauss' law to obtain the expression for the electric…2023 · asked 3×
- Two small solid metal balls A and B of radii R and 2 R having charge densities 2 σ and 3 σ respectively are…2025 · asked 3×
- Four point charges of 1 μ C,-2 μ C, 1 μ C and -2 μ C are placed at the corners A, B, C and D respectively, of…2024 · asked 3×
- (i) Use Gauss' law to obtain an expression for the electric field due to an infinitely long thin straight…2023 · asked 3×
- Which one of the following statements is correct? Electric field due to static charges is2025 · asked 3×
- A point charge situated at a distance ' r ' from a short electric dipole on its axis, experiences a force F.…2023 · asked 3×
CBSE Class 12 Physics past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.