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Physics · 2025 · 5 marks
CBSE 2025 · Region 6 · Set 2 · Q32
(a) (i) The electric field in a region is given by E =40 x i N / C. Find the amount of work done in taking a unit positive charge from a point (0,3…
(a)(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.
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(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
\[\phi_L = \int \vec{E}.\vec{ds}\]
\[= -E_L(L^2)\]
\[= -5L^2\]
\[\phi_R = E_R(L^2)\]
\[= (10L+5)L^2\]
\[\phi_{net} = \phi_L + \phi_R\]
\[= -5L^2 + (10L+5)L^2\]
\[= 10L^3\ \text{Nm}^2/\text{C}\]
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