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Physics · 2026 · 5 marks
CBSE 2026 · Region 4 · Set 1 · Q33
(a) (i) Explain the following statements giving reason: (I) An equipotential surface through a point is normal to the electric field at that point.…
(i)
Explain the following statements giving reason :
(I)
An equipotential surface through a point is normal to the electric field at that point.
(II)
When a dielectric is placed in an external electric field, the electric field inside the dielectric is less than that outside it.
(III)
The potential difference between the plates of a charged parallel plate capacitor decreases when its plates are brought closer.
(ii)
Obtain an expression for the work done to dissociate the system of three charges q, - 4q and 2q placed at the vertices A, B and C respectively of an equilateral triangle of side 'a'.
(i)
Answer the following giving reason :
(I)
The electron drift speed is estimated to be only a few mm/s for currents in the range of a few amperes. How, then, is the current established almost the instant a circuit is closed ?
(II)
A low voltage supply from which one needs high currents must have very low internal resistance. Why ?
(III)
The assertion that V = IR is a statement of Ohm's law is not true. Why ?
(ii)
Two cells of emfs $\displaystyle 12$ V and $\displaystyle 6$ V are connected in parallel as shown in the figure. Their internal resistances are $\displaystyle 1 \Omega$ and $\displaystyle 0.5 \Omega$ respectively. Calculate the emf and internal resistance of the equivalent cell between points A and B.
Marking-scheme solution
(a)(i)
Reason for:
(I)
Electric field perpendicular to the equipotential surface at a point. $\displaystyle 1$
(II)
Reduction of electric field inside the dielectric. $\displaystyle 1$
(III)
Potential difference decreases when the plates of a capacitor are brought closer. $\displaystyle 1$
(ii)
Obtaining expression for the work done.
(i)
(I)
If the field were not normal to the equipotential surface, it would have non-zero component along the surface. To move a unit test charge against the direction of the component of the field, work would be done. This is in contradiction to the definition of equipotential surface.
Alternatively
$\displaystyle \oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dl}}=\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}} ; \mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}}$ for equipotential surface
$\displaystyle \oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dl}}=0 \Rightarrow \cos \theta=0$ as $\displaystyle \mathrm{E} \neq 0 \& \mathrm{dl} \neq 0$ and $\displaystyle \theta=90^{0}$
(II)
When a dielectric is placed in an external electric field, due to polarization of its molecules, an electric field is induced in a direction opposite to the external field, hence net electric field inside is reduced.
Alternatively
\[\varepsilon=\varepsilon_{0}-\varepsilon_{\mathrm{P}}
\]
(III)
Potential difference (V) $\displaystyle =\frac{\mathrm{Q}}{\mathrm{C}}$
\[\begin{aligned}
& \mathrm{V}=\frac{\mathrm{Qd}}{\mathrm{~A} \varepsilon_{0}} \\
& \mathrm{~V} \propto \mathrm{~d}
\end{aligned}
\]
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CBSE Class 12 Physics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.