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CBSE 2026 · Region 5 · Set 1 · Q29 · 4 marks

The electric potential (V) and electric field (E) are closely related concepts in electrostatics. The electric field is a vector quantity that represents the force per unit charge at a given point in space, whereas electric potential is a scalar quantity that represents the potential energy per unit charge at a given point in space. Electric field and electric potential are related by the equations $\displaystyle \mathrm{E}_{\mathrm{r}}=\frac{-\mathrm{dV}}{\mathrm{dr}}$ and $\displaystyle \overrightarrow{\mathrm{E}}=\mathrm{E}_{\mathrm{r}} \hat{\mathrm{r}}$, i.e., electric field is the negative gradient of the electric potential. This means that electric field points in the direction of decreasing potential and its magnitude is the rate of change of potential with distance. The electric field is the force that drives a unit charge to move from higher potential region to lower potential region and electric potential difference between the two points determines the work done in moving a unit charge from one point to the other point.
Figure: CBSE Class 12 Physics 2026, Electrostatic Potential and Capacitance
A pair of square conducting plates having sides of length $\displaystyle 0.05$ m are arranged parallel to each other in x-y plane. They are $\displaystyle 0.01$ m apart along z-axis and are connected to a $\displaystyle 200$ V power supply as shown in the figure. An electron enters with a speed of $\displaystyle 3 \times 10^{7} \mathrm{~ms}^{-1}$ horizontally and symmetrically in the space between the two plates. Neglect the effect of gravity on the electron.
(i)
The electric field $\displaystyle \overrightarrow{\mathrm{E}}$ in the region between the plates is :
(A)
$\displaystyle \left(2 \times 10^{2} \frac{\mathrm{~V}}{\mathrm{~m}}\right) \hat{\mathrm{k}}$
(B)
$\displaystyle -\left(2 \times 10^{2} \frac{\mathrm{~V}}{\mathrm{~m}}\right) \hat{\mathrm{k}}$
(C)
$\displaystyle \left(2 \times 10^{4} \frac{\mathrm{~V}}{\mathrm{~m}}\right) \hat{\mathrm{k}}$
(D)
$\displaystyle -\left(2 \times 10^{4} \frac{\mathrm{~V}}{\mathrm{~m}}\right) \hat{\mathrm{k}}$
(ii)
In the region between the plates, the electron moves with an acceleration $\displaystyle \overrightarrow{\mathrm{a}}$ given by :
(A)
$\displaystyle -\left(3 \cdot 5 \times 10^{15} \mathrm{~ms}^{-2}\right) \hat{\mathrm{k}}$
(B)
$\displaystyle \left(3.5 \times 10^{15} \mathrm{~ms}^{-2}\right) \hat{\mathrm{k}}$
(C)
$\displaystyle \left(3 \cdot 5 \times 10^{13} \mathrm{~ms}^{-2}\right) \hat{\mathrm{i}}$
(D)
$\displaystyle -\left(3 \cdot 5 \times 10^{13} \mathrm{~ms}^{-2}\right) \hat{\mathrm{i}}$
(iii)
Time interval during which an electron moves through the region between the plates is :
(A)
$\displaystyle 9.0 \times 10^{-9} \mathrm{~s}$
(B)
$\displaystyle 1.67 \times 10^{-8} \mathrm{~s}$
(C)
$\displaystyle 1.67 \times 10^{-9} \mathrm{~s}$
(D)
$\displaystyle 2 \cdot 17 \times 10^{-9} \mathrm{~s}$
(iv)
Which one of the following is the path traced by the electron in between the two plates ?
(A)
a (B) b (C) c (D) d

Electrostatic Potential and CapacitanceThe Parallel Plate CapacitorApplycase_studyhard

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CBSE Class 12 Physics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.

The electric potential (V) and electric field (E) are closely… — CBSE Class 12 Physics 2026 | SolveIt