CBSE 2025 · Region 2 · Set 1 · Q31 · 5 marks
(i)A small conducting sphere A of radius r charged to a potential V , is enclosed by a spherical conducting shell B of radius R . If A and $\displaystyle B$ are connected by a thin wire, calculate the final potential on sphere $\displaystyle A$ and shell $\displaystyle B$.(ii)Write two characteristics of equipotential surfaces. A uniform electric field of $\displaystyle 50 \mathrm{NC}^{-1}$ is set up in a region along $\displaystyle +x$ axis. If the potential at the origin $\displaystyle ( 0,0 )$ is $\displaystyle 220$ V , find the potential at a point ( $\displaystyle 4 \mathrm{~m}, 3 \mathrm{~m}$ ).(i)What is difference between an open surface and a closed surface ? Draw elementary surface vector $\displaystyle d \vec{\mathrm{S}}$ for a spherical surface $\displaystyle \mathrm{S}$.(ii)Define electric flux through a surface. Give the significance of a Gaussian surface. A charge outside a Gaussian surface does not contribute to total electric flux through the surface. Why ?(iii)A small spherical shell $\displaystyle \mathrm{S}_{1}$ has point charges $\displaystyle \mathrm{q}_{1}=-3 \mu \mathrm{C}, \mathrm{q}_{2}=-2 \mu \mathrm{C}$ and $\displaystyle \mathrm{q}_{3}=9 \mu \mathrm{C}$ inside it. This shell is enclosed by another big spherical shell $\displaystyle \mathrm{S}_{2}$. A point charge $\displaystyle Q$ is placed in between the two surfaces $\displaystyle \mathrm{S}_{1}$ and $\displaystyle \mathrm{S}_{2}$. If the electric flux through the surface $\displaystyle \mathrm{S}_{2}$ is four times the flux through surface $\displaystyle \mathrm{S}_{1}$, find charge Q .
(i)
A small conducting sphere A of radius r charged to a potential V , is enclosed by a spherical conducting shell B of radius R . If A and $\displaystyle B$ are connected by a thin wire, calculate the final potential on sphere $\displaystyle A$ and shell $\displaystyle B$.
(ii)
Write two characteristics of equipotential surfaces. A uniform electric field of $\displaystyle 50 \mathrm{NC}^{-1}$ is set up in a region along $\displaystyle +x$ axis. If the potential at the origin $\displaystyle ( 0,0 )$ is $\displaystyle 220$ V , find the potential at a point ( $\displaystyle 4 \mathrm{~m}, 3 \mathrm{~m}$ ).
(i)
What is difference between an open surface and a closed surface ? Draw elementary surface vector $\displaystyle d \vec{\mathrm{S}}$ for a spherical surface $\displaystyle \mathrm{S}$.
(ii)
Define electric flux through a surface. Give the significance of a Gaussian surface. A charge outside a Gaussian surface does not contribute to total electric flux through the surface. Why ?
(iii)
A small spherical shell $\displaystyle \mathrm{S}_{1}$ has point charges $\displaystyle \mathrm{q}_{1}=-3 \mu \mathrm{C}, \mathrm{q}_{2}=-2 \mu \mathrm{C}$ and $\displaystyle \mathrm{q}_{3}=9 \mu \mathrm{C}$ inside it. This shell is enclosed by another big spherical shell $\displaystyle \mathrm{S}_{2}$. A point charge $\displaystyle Q$ is placed in between the two surfaces $\displaystyle \mathrm{S}_{1}$ and $\displaystyle \mathrm{S}_{2}$. If the electric flux through the surface $\displaystyle \mathrm{S}_{2}$ is four times the flux through surface $\displaystyle \mathrm{S}_{1}$, find charge Q .
Marking-scheme solution
(i)
Potential on sphere A $\displaystyle =V=\frac{Q}{4 \pi \varepsilon_{0} r}$
Charge on sphere A $\displaystyle =4 \pi \varepsilon_{0} r V$
The charge is transferred to shell B.
Potential on shell B $\displaystyle =\frac{1}{4 \pi \varepsilon_{0}} \times \frac{4 \pi \varepsilon_{0} r V}{R}$
Potential on shell B $\displaystyle =\frac{r V}{R}$
Potential on sphere A = Potential on shell B
(ii)
Characteristics of equipotential surfaces:
Potential at all points on the surface is same.
Equipotential surface is normal to the direction of the electric field.
The work done in moving a charge on an equipotential surface is zero.
$\displaystyle V_{0}-V=E d=50 \times 4$
$\displaystyle V_{0}-V=200 \mathrm{~V}$
$\displaystyle V=220 \mathrm{~V}-200 \mathrm{~V}$
$\displaystyle V=20 \mathrm{~V}$
(i)
Open Surface – A surface which does not enclose a volume.
Closed Surface – A surface which does enclose a volume.
(ii)
Electric flux is defined as the number of electric field lines crossing an area normally.
$\displaystyle \phi=\vec{E} \cdot \vec{A}$
$\displaystyle \phi=E A \cos \theta$
Significance of Gaussian Surface:
It helps in finding the electric field in a simpler way.
Reason:
Because any electric field line from the charge which enters the surface at one point will exit at another, resulting in a net zero flux.
(iii)
Total charge enclosed by $\displaystyle S_{1}=(-3-2+9) \mu \mathrm{C}=4 \mu \mathrm{C}$
Total charge enclosed by $\displaystyle S_{2}=Q+4 \mu \mathrm{C}$
$\displaystyle \phi_{S_{2}}=4 \phi_{S_{1}}$
$\displaystyle \frac{Q+4 \mu \mathrm{C}}{\varepsilon_{0}}=4\left(\frac{4 \mu \mathrm{C}}{\varepsilon_{0}}\right)$
$\displaystyle Q=12 \mu \mathrm{C}$
Electric Charges and FieldsElectric FluxApplylong_answerhard
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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.