CHAPTER 2: ELECTROSTATIC POTENTIAL AND CAPACITANCE
SAMPLE QUESTION PAPER (ASSIGNMENT-1)
- There are 33 questions in all. All questions are compulsory.
- Section A: 16 questions (12 MCQs and 4 Assertion-Reasoning) of 1 mark each.
- Section B: 5 short answer questions of 2 marks each.
- Section C: 7 short answer questions of 3 marks each.
- Section D: 2 case study-based questions of 4 marks each.
- Section E: 3 long answer questions of 5 marks each.
SECTION A (16 Marks)
Q.1. Equipotential surfaces:
Explanation: Since $\displaystyle {E = -{\frac{{dV}}{{dr}}} \implies dr = -{\frac{{dV}}{{E}}}}$, for constant $\displaystyle {dV}$, distance $\displaystyle {dr}$ is smaller where $\displaystyle {E}$ is larger.
Q.2. Two particles A and B of same mass but having charges $\displaystyle {q}$ and $\displaystyle {4q}$ respectively are accelerated from rest through different potential differences $\displaystyle {V_A}$ and $\displaystyle {V_B}$ such that they attain the same kinetic energy. The ratio $\displaystyle {V_A : V_B}$ is:
Explanation: $\displaystyle {K = q V \implies q_A V_A = q_B V_B \implies q V_A = 4q V_B \implies {\frac{{V_A}}{{V_B}}} = 4}$.
Q.3. At what rate does the electric field change between the plates of a square capacitor of side $\displaystyle {5\text{ cm}}$ if the plates are spaced $\displaystyle {1.2\text{ mm}}$ apart and the voltage across them is changing at the rate of $\displaystyle {60\text{ V/s}}$?
Explanation: $\displaystyle {E = {\frac{{V}}{{d}}} \implies {\frac{{dE}}{{dt}}} = {\frac{{1}}{{d}}}{\frac{{dV}}{{dt}}} = {\frac{{60}}{{1.2 \times 10^{-3}}}} = 5 \times 10^4\text{ Vm}^{-1}\text{s}^{-1}}$.
Q.4. When a dielectric material is introduced between the plates of a charged isolated condenser, the electric field between the plates:
Explanation: $\displaystyle {E = {\frac{{E_0}}{{K}} stroke}}$. Since $\displaystyle {K > 1}$, the induced field opposes the applied field, reducing the net electric field.
Q.5. In a certain region of space with volume $\displaystyle {0.2\text{ m}^3}$, the electric potential is found to be $\displaystyle {5\text{ V}}$ throughout. The magnitude of electric field in this region is:
Explanation: $\displaystyle {E = -{\frac{{dV}}{{dr}}}$. Since potential $V$ is constant throughout the volume, $dV = 0 \implies E = 0$.
Q.6. A capacitor of capacitance $\displaystyle {C}$ has charge $\displaystyle {Q}$ and energy $\displaystyle {W}$ stored. If charge is increased up to $\displaystyle {2Q}$, the energy stored will be:
Explanation: $\displaystyle {W = {\frac{{Q^2}}{{2C}}}$. When charge doubles to $2Q$, $\displaystyle {W' = {\frac{{(2Q)^2}}{{2C}}} = 4 {\left( {\frac{{Q^2}}{{2C}}} \right)} = 4W}$.
Q.7. A capacitor is charged by a battery. After disconnecting the battery, the plates are pulled apart. The potential difference between plates will:
Explanation: Charge $\displaystyle {Q}$ remains constant. As distance $\displaystyle {d}$ increases, capacitance $\displaystyle {C = {\frac{{\varepsilon_0 A}}{{d}}}}$ decreases. Hence $\displaystyle {V = {\frac{{Q}}{{C}}}}$ increases.
Q.8. A $\displaystyle {10\,\mu\text{F}}$ capacitor is connected in series with a $\displaystyle {20\,\mu\text{F}}$ capacitor across a $\displaystyle {30\text{ V}}$ supply. The charge on each capacitor is:
Explanation: $\displaystyle {C_{\text{eq}} = {\frac{{10 \times 20}}{{10 + 20}}} = {\frac{{200}}{{30}}} = {\frac{{20}}{{3}}}\,\mu\text{F}}$. Charge $\displaystyle {Q = C_{\text{eq}} V = {\left( {\frac{{20}}{{3}}} \right)} \times 30 = 200\,\mu\text{C}}$.
Q.9. A force of $\displaystyle {10\text{ N}}$ acts on a charged particle placed between two plates of a charged capacitor. If one plate of the capacitor is removed, the force acting on that particle will be:
Explanation: Electric field between two plates is $\displaystyle {E = {\frac{{\sigma}}{{\varepsilon_0}}}}$. One plate contributes $\displaystyle {{\frac{{\sigma}}{{2\varepsilon_0}}} = {\frac{{E}}{{2}}}}$. Halving field halves the force to $5\text{ N}$.
Q.10. An electron with kinetic energy $\displaystyle {K_1}$ enters between plates of a capacitor at an angle $\displaystyle {\alpha}$ with the plates. It leaves the plate at an angle $\displaystyle {\beta}$ with kinetic energy $\displaystyle {K_2}$. The ratio $\displaystyle {K_1 : K_2}$ is:
Explanation: Horizontal velocity component remains constant: $\displaystyle {v_1 \cos\alpha = v_2 \cos\beta \implies {\frac{{v_1^2}}{{v_2^2}}} = {\frac{{\cos^2\beta}}{{\cos^2\alpha}}} \implies {\frac{{K_1}}{{K_2}}} = {\frac{{\cos^2\beta}}{{\cos^2\alpha}}}}$.
Q.11. A parallel plate capacitor filled with a medium of dielectric constant 10 is connected across a battery and charged. The dielectric slab is replaced by another slab of dielectric constant 15, then energy of the capacitor will:
Explanation: Voltage $V$ is constant. $\displaystyle {U = {\frac{{1}}{{2}}} C V^2 \propto K}$. Ratio $\displaystyle {{\frac{{U_2}}{{U_1}}} = {\frac{{15}}{{10}}} = 1.5 \implies 50\%}$ increase.
Q.12. A hydrogen ion ($\displaystyle {\text{H}^+}$) and a singly ionized helium atom ($\displaystyle {\text{He}^+}$) are accelerated from rest through the same potential difference. The ratio of final speed of hydrogen and helium ions is close to:
Explanation: $\displaystyle {q V = {\frac{{1}}{{2}}} m v^2 \implies v = \sqrt{{\frac{{2 q V}}{{m}}}} \propto {\frac{{1}}{{\sqrt{m}}}}}$. Since $\displaystyle {m_{\text{He}} \approx 4 m_{\text{H}}}$, ratio $\displaystyle {{\frac{{v_{\text{H}}}{{v_{\text{He}}}}} = \sqrt{4} = 2:1}$.
Directions for Q.13 to Q.16:
(A) Both Assertion and Reason are true and Reason is correct explanation.
(B) Both Assertion and Reason are true but Reason is NOT correct explanation.
(C) Assertion is true but Reason is false.
(D) Both Assertion and Reason are false.
Q.13. Assertion (A): Equipotential surfaces are always perpendicular to electric field lines.
Reason (R): No work is done in moving a test charge along an equipotential surface.
Explanation: $\displaystyle {dW = \vec{E} \cdot d\vec{r} = E dr \cos\theta = 0 \implies \cos\theta = 0 \implies \theta = 90^\circ}$.
Q.14. Assertion (A): When two capacitors are connected in series, total capacitance is less than the smaller capacitance.
Reason (R): In series connection, potential difference across each capacitor is different.
Explanation: Formula $\displaystyle {{\frac{{1}}{{C_{\text{eq}}}}} = {\frac{{1}}{{C_1}}} + {\frac{{1}}{{C_2}}}}$ mathematical proves $\displaystyle {C_{\text{eq}} < C_1, C_2}$, not potential division.
Q.15. Assertion (A): An electron has a higher potential energy when at a location associated with a more negative value of potential and a low potential energy when at a location associated with a more positive potential.
Reason (R): An electron moves from a region of higher potential to region of lower potential.
Explanation: $\displaystyle {U = q V = -e V}$, so negative potential gives positive (higher) potential energy. Electrons naturally move from lower to higher potential.
Q.16. Assertion (A): If a dielectric is placed in external field, field inside dielectric will be less than applied field.
Reason (R): Electric field will induce dipole moment opposite to field direction.
Explanation: Induced dipole moment aligns in the direction of applied field, creating an internal induced electric field opposite to the external field.
SECTION B (10 Marks)
Q.17. Two charged spherical conductors of radius $\displaystyle {R_1}$ and $\displaystyle {R_2}$ when connected by a conducting wire acquire charges $\displaystyle {Q_1}$ and $\displaystyle {Q_2}$ respectively. Find the ratio of their surface charge densities in terms of their radii.
Surface charge density ratio: $\displaystyle {{\frac{{\sigma_1}}{{\sigma_2}}} = {\frac{{Q_1 / 4\pi R_1^2}}{{Q_2 / 4\pi R_2^2}}} = {\left( {\frac{{Q_1}}{{Q_2}}} \right)} {\left( {\frac{{R_2^2}}{{R_1^2}}} \right)} = {\left( {\frac{{R_1}}{{R_2}}} \right)} {\left( {\frac{{R_2^2}}{{R_1^2}}} \right)} = {\frac{{R_2}}{{R_1}}}}$.
Q.18. Derive the expression for the electric potential due to an electric dipole at a point on its axial line.
$\displaystyle {V = V_{+q} + V_{-q} = {\frac{{1}}{{4\pi\varepsilon_0}}} {\frac{{q}}{{r-a}}} - {\frac{{1}}{{4\pi\varepsilon_0}}} {\frac{{q}}{{r+a}}} = {\frac{{q}}{{4\pi\varepsilon_0}}} {\left[ {\frac{{(r+a)-(r-a)}}{{r^2-a^2}}} \right]} = {\frac{{1}}{{4\pi\varepsilon_0}}} {\frac{{p}}{{r^2-a^2}}}}$.
For short dipole ($\displaystyle {r \gg a}$), $\displaystyle {V = {\frac{{1}}{{4\pi\varepsilon_0}}} {\frac{{p}}{{r^2}}}}$.
Q.19. The two graphs show variations of electrostatic potential ($\displaystyle {V}$) with $\displaystyle {1/r}$ ($\displaystyle {r}$ being distance from point charge) for two-point charges $\displaystyle {q_1}$ and $\displaystyle {q_2}$.
(i) What are the signs of the two charges?
(ii) Which of the two charges has the larger magnitude and why?
(i) $\displaystyle {q_1}$ is positive (slope $\displaystyle {V / (1/r) > 0}$) and $\displaystyle {q_2}$ is negative (slope $< 0$).
(ii) Magnitude of slope equals $\displaystyle {k |q|}$. Since line for $\displaystyle {q_2}$ has steeper magnitude of slope than $\displaystyle {q_1}$, $\displaystyle {|q_2| > |q_1|}$.
Q.20. Two capacitors of different capacitance are connected first in series and then in parallel across a DC source of $\displaystyle {100\text{ V}}$. If total energy stored in combination in two cases are $\displaystyle {40\text{ mJ}}$ and $\displaystyle {250\text{ mJ}}$ respectively, find the capacitances.
You have N similar capacitors each of capacitance $\displaystyle {Y\text{ farad}}$. How will you connect them to obtain maximum & minimum equivalent capacitance? Ratio of maximum to minimum equivalent capacitance?
Parallel: $\displaystyle {U_p = {\frac{{1}}{{2}}} (C_1 + C_2) V^2 = 250 \times 10^{-3} \implies C_1 + C_2 = {\frac{{2 \times 0.250}}{{10000}}} = 50\,\mu\text{F}}$.
Series: $\displaystyle {U_s = {\frac{{1}}{{2}}} {\left( {\frac{{C_1 C_2}}{{C_1 + C_2}}} \right)} V^2 = 40 \times 10^{-3} \implies {\frac{{C_1 C_2}}{{50}}} = {\frac{{2 \times 0.040}}{{10000}}} \implies C_1 C_2 = 400}$.
Solving gives $\displaystyle {C_1 = 40\,\mu\text{F}}$ and $\displaystyle {C_2 = 10\,\mu\text{F}}$.
OR Part: Max capacitance in parallel ($\displaystyle {C_{\text{max}} = N Y}$), Min capacitance in series ($\displaystyle {C_{\text{min}} = Y / N}$). Ratio $\displaystyle {C_{\text{max}} / C_{\text{min}} = N^2}$.
Q.21. An infinite plane sheet of charge density $\displaystyle {10^{-8}\text{ C/m}^2}$ held in air. How far apart are equipotential surfaces whose potential difference is $\displaystyle {5\text{ V}}$?
A capacitor is connected across a battery:
I) Why does each plate receive a charge of exactly same magnitude?
II) Is this true even if plates are of different sizes?
Distance $\displaystyle {\Delta r = {\frac{{\Delta V}}{{E}}} = {\frac{{5}}{{564.7}}} = 8.85 \times 10^{-3}\text{ m} = 8.85\text{ mm}}$.
OR Part: I) Due to conservation of charge and electrostatic induction, charge flowing from one terminal equals charge arriving at the other. II) Yes, electrostatic induction ensures equal and opposite charges regardless of plate dimensions.
SECTION C (21 Marks)
Q.22. A regular hexagon of side $\displaystyle {10\text{ cm}}$ has charge $\displaystyle {5\,\mu\text{C}}$ at each of its vertices. What is the resultant potential at the centre of the hexagon?
$\displaystyle {V_{\text{net}} = 6 \times {\left( {\frac{{1}}{{4\pi\varepsilon_0}}} {\frac{{q}}{{r}}} \right)} = 6 \times {\frac{{9 \times 10^9 \times 5 \times 10^{-6}}}{{0.1}}} = 2.7 \times 10^6\text{ V}}$.
Q.23. Consider two identical positive point charges located at points $\displaystyle {(0,0)}$ and $\displaystyle {(a,0)}$.
(i) Is there a point on the line joining them at which the electric field is zero?
(ii) Is there a point on the line joining them at which the electric potential is zero?
(i) Yes, at midpoint $\displaystyle {(a/2, 0)}$, electric fields due to both positive charges are equal in magnitude and opposite in direction, cancelling to zero.
(ii) No, since potential is scalar and both charges are positive, $\displaystyle {V = V_1 + V_2 > 0}$ everywhere on finite distances.
Q.24. A parallel plate capacitor A of capacitance $\displaystyle {C}$ is charged by a battery to voltage $\displaystyle {V}$. The battery is disconnected and an uncharged capacitor B of capacitance $\displaystyle {2C}$ is connected across A. Find ratio of:
i) Final charges on A and B
ii) Total electrostatic energy stored in A and B finally vs initially.
i) $\displaystyle {Q_A' = C {\left( {\frac{{V}}{{3}}} \right)} = {\frac{{CV}}{{3}}}}$, $\displaystyle {Q_B' = 2C {\left( {\frac{{V}}{{3}}} \right)} = {\frac{{2CV}}{{3}}} \implies {\frac{{Q_A'}}{{Q_B'}}} = 1:2}$.
ii) Initial energy $\displaystyle {U_i = {\frac{{1}}{{2}}} C V^2}$. Final energy $\displaystyle {U_f = {\frac{{1}}{{2}}} (3C) {\left( {\frac{{V}}{{3}}} \right)}^2 = {\frac{{1}}{{6}}} C V^2 \implies {\frac{{U_f}}{{U_i}}} = 1:3}$.
Q.25. Define an equipotential surface.
(A) Draw equipotential surfaces (i) for a single point charge, (ii) in a constant electric field in z-direction.
(B) Why are equipotential surfaces about a single charge not equidistant?
(C) Can electric field exist tangential to an equipotential surface? Justify.
(B) Since $\displaystyle {E \propto 1/r^2}$, field decreases with distance, so for equal potential steps $\displaystyle {\Delta V}$, spacing $\displaystyle {\Delta r = \Delta V / E}$ increases.
(C) No, if tangential field existed, work would be done moving charge along the surface, violating $\displaystyle {\Delta V = 0}$.
Q.26. If N drops of same size each having the same charge coalesce to form a bigger drop, how will the following vary with respect to single small drop:
i) Total charge, ii) Potential, iii) Capacitance?
Two conducting spherical shells A and B of radii R and 2R are kept far apart and charged to the same surface charge density $\displaystyle {\sigma}$. They are connected by a wire. Obtain expression for final potential of shell A.
i) $\displaystyle {Q = N q}$.
ii) $\displaystyle {V' = {\frac{{k Q}}{{R}}} = {\frac{{k N q}}{{N^{1/3} r}}} = N^{2/3} V}$.
iii) $\displaystyle {C' = 4\pi\varepsilon_0 R = N^{1/3} C}$.
OR Part: Initial charges $\displaystyle {Q_A = \sigma(4\pi R^2)}$, $\displaystyle {Q_B = \sigma(4\pi (2R)^2) = 4 Q_A}$. Total charge $\displaystyle {Q = 5 \sigma (4\pi R^2)}$.
Final common potential $\displaystyle {V = {\frac{{Q_{\text{total}}}{{C_A + C_B}}} = {\frac{{5\sigma (4\pi R^2)}}{{4\pi\varepsilon_0 R + 4\pi\varepsilon_0 (2R)}} = {\frac{{5\sigma R}}{{3\varepsilon_0}}}}$.
Q.27. X and Y are two equipotential surfaces separated by $2\text{ m}$ in uniform electric field $10\text{ V/m}$. Potential of surface X is $10\text{ V}$.
i) Find potential of surface Y.
ii) Work done in moving $+2\text{ C}$ charge from surface Y to surface X along path 1 and path 2?
i) Potential decreases along field lines ($\displaystyle {\Delta V = E d = 10 \times 2 = 20\text{ V}}$). Potential at Y is $\displaystyle {10 + 20 = 30\text{ V}}$.
ii) $\displaystyle {W = q (V_X - V_Y) = 2 \times (10 - 30) = -40\text{ J}}$. Work done along path 2 is also $\displaystyle {-40\text{ J}}$ because electrostatic field is conservative.
Q.28. In a parallel plate capacitor with air between plates, area is $\displaystyle {0.0006\text{ m}^2}$ and separation is $\displaystyle {3\text{ mm}}$.
A) Calculate capacitance.
B) If connected to $100\text{ V}$ supply, charge on each plate?
C) How affected if $3\text{ mm}$ thick mica sheet ($K=6$) is inserted while supply remains connected?
A) $\displaystyle {C = {\frac{{\varepsilon_0 A}}{{d}}} = {\frac{{8.854 \times 10^{-12} \times 0.0006}}{{3 \times 10^{-3}}}} = 1.77 \times 10^{-11}\text{ F} = 17.7\text{ pF}}$.
B) $\displaystyle {Q = C V = 1.77 \times 10^{-11} \times 100 = 1.77 \times 10^{-9}\text{ C} = 1.77\text{ nC}}$.
C) Voltage stays $100\text{ V}$. New charge $\displaystyle {Q' = K Q = 6 \times 1.77\text{ nC} = 10.62\text{ nC}}$.
SECTION D: CASE STUDY QUESTIONS (8 Marks)
Q.29. Case Study 1: A parallel plate capacitor consists of two parallel metal plates X and Y of area A separated by distance d with surface charge density $\displaystyle {\sigma}$.
(i) Potential difference between plates of parallel plate capacitor is given by:
(a) $\displaystyle {Q d / (\varepsilon_0 A)}$ (b) $\displaystyle {d \varepsilon_0 / A Q}$ (c) $\displaystyle {A d / (\varepsilon_0 Q)}$ (d) $\displaystyle {Q A / d \varepsilon_0}$
(ii) Battery disconnected and dielectric slab inserted. Charge on capacitor:
(a) Remains constant (b) Increases (c) Decreases (d) Zero
(iii) Capacitance does not depend on:
(a) Area (b) Type of metal (c) Separation distance (d) Medium
(iv) Parallel plate capacitor has $\displaystyle {10\,\mu\text{F}}$. If distance doubled, new capacitance is:
(a) $\displaystyle {20\,\mu\text{F}}$ (b) $\displaystyle {15\,\mu\text{F}}$ (c) $\displaystyle {10\,\mu\text{F}}$ (d) $\displaystyle {5\,\mu\text{F}}$
Q.30. Case Study 2: RC Circuit Charging & Discharging.
(a) Draw graph of variation of voltage across resistor during charging.
(b) At time T (fully charged), what is current in circuit?
(c) Switch moved to position 2: how does voltage across resistor and capacitor change?
(d) Dielectric inserted: effect on (i) capacitance, (ii) charge when switch in position 1?
(a) Exponentially decaying curve starting from $\displaystyle {V_0}$ down to 0.
(b) Current is zero because capacitor is fully charged ($\displaystyle {V_C = V_0}$).
(c) Both decay exponentially to zero as capacitor discharges through resistor.
(d) (i) Capacitance increases $\displaystyle {C' = K C}$, (ii) Charge increases $\displaystyle {Q' = K Q}$.
SECTION E (15 Marks)
Q.31. (a) Sketch equipotential surface for (i) $\displaystyle {Q_1 + Q_2 = 0}$ (dipole), (ii) Two identical positive charges.
(b) Write 3 characteristics of equipotential surface.
(c) Potential in region varies as $\displaystyle {V = 5 + 4x^2}$. i) Is electric field uniform? ii) Force on $\displaystyle {1\text{ C}}$ at $\displaystyle {x = -1\text{ m}}$.
(a) Expression for capacitance of parallel plate capacitor with dielectric slab of thickness $\displaystyle {t < d}$.
(b) Copper sphere in ocean: capacitance increase, decrease or same?
(c) Plot V vs d graph for capacitor connected to potential source V as d increases.
(c) $\displaystyle {E = -{\frac{{dV}}{{dx}}} = -8x\text{ N/C}}$. i) Non-uniform (depends on x). ii) At $\displaystyle {x = -1\text{ m}}$, $\displaystyle {E = +8\text{ N/C} \implies F = q E = 1 \times 8 = 8\text{ N}}$ along +x direction.
OR Part: (a) $\displaystyle {C = {\frac{{\varepsilon_0 A}}{{d - t + t/K}}}}$. (b) Increase, because surrounding conductive saltwater acts as outer spherical conductor of infinite radius. (c) $V$ remains constant straight horizontal line.
Q.32. (a) Parallel plate capacitor filled with slab $K$ versus filled with two slabs $K_1, K_2$ of thickness $d/2$. Prove $\displaystyle {K = {\frac{{2 K_1 K_2}}{{K_1 + K_2}}}}$.
(b) Why no steady state current in capacitor across DC battery?
(a) Derive potential due to point charge $\displaystyle {V = {\frac{{1}}{{4\pi\varepsilon_0}}}{\frac{{q}}{{r}}}}$. Spherical symmetry?
(b) Work done moving $20\text{ mC}$ from $M(6,0,0)$ to $N(-4,0,0)$ in dipole field at $(0,0,-a)$ and $(0,0,a)$.
(c) What is electrostatic shielding?
(a) Series combination: $\displaystyle {{\frac{{1}}{{C}}} = {\frac{{1}}{{C_1}}} + {\frac{{1}}{{C_2}}} \implies {\frac{{d}}{{K \varepsilon_0 A}}} = {\frac{{d/2}}{{K_1 \varepsilon_0 A}}} + {\frac{{d/2}}{{K_2 \varepsilon_0 A}}} \implies {\frac{{1}}{{K}}} = {\frac{{1}}{{2}}} {\left( {\frac{{1}}{{K_1}} + {\frac{{1}}{{2 K_2}}} \right)} \implies K = {\frac{{2 K_1 K_2}}{{K_1 + K_2}}}}$.
(b) In steady state, capacitor potential equals battery voltage, opposing further charge flow.
OR Part: (b) Points M and N lie on equatorial xy-plane where potential is zero everywhere. $\displaystyle {W = q (V_N - V_M) = 0}$.
Q.33. (a) Derive expression for capacitance of parallel plate capacitor in air.
(b) Obtain equivalent capacitance and charge on each capacitor in network connected to $300\text{ V}$ supply.
Series with $\displaystyle {C_4}$: $\displaystyle {C_{\text{eq}} = {\frac{{200 \times 100}}{{200 + 100}}} = {\frac{{200}}{{3}}}\text{ pF}}$.
Charge $\displaystyle {Q_4 = {\left( {\frac{{200}}{{3}}} \right)} \times 300 = 2 \times 10^{-8}\text{ C}}$.
ASSIGNMENT – 2 (MCQ PRACTICE)
1. The electric potential at a point 'P' due to a point charge of $\displaystyle {5 \times 10^{-9}\text{ C}}$ is $\displaystyle {50\text{ V}}$. The distance of P from the point charge is:
Explanation: $\displaystyle {V = {\frac{{k q}}{{r}}} \implies 50 = {\frac{{9 \times 10^9 \times 5 \times 10^{-9}}}{{r}}} \implies r = {\frac{{45}}{{50}}} = 0.9\text{ m} = 90\text{ cm}}$.
2. Electrostatic potential due to an electric dipole at distance 'r' varies as:
3. 64 identical drops each charged up to potential of $\displaystyle {10\text{ mV}}$ are combined to form a bigger drop. Potential of bigger drop is:
Explanation: $\displaystyle {V' = N^{2/3} V = (64)^{2/3} \times 10 = 16 \times 10 = 160\text{ mV}}$.
4. A spherical drop of mercury having potential $\displaystyle {2.5\text{ V}}$ is obtained as a result of merging 125 droplets. Potential of constituent droplet is:
Explanation: $\displaystyle {V' = N^{2/3} v \implies 2.5 = (125)^{2/3} v = 25 v \implies v = 0.1\text{ V}}$.
5. Three point charges each of charge $\displaystyle {q}$ placed on vertices of triangle ABC with AB = AC = 5L, BC = 6L. Electrostatic potential at midpoint of BC is:
Explanation: Altitude from A to BC is $\displaystyle {\sqrt{(5L)^2 - (3L)^2} = 4L}$. Distance to B and C midpoints is $3L$. $\displaystyle {{V = {\frac{{kq}}{{4L}}} + {\frac{{kq}}{{3L}}} + {\frac{{kq}}{{3L}}} = kq{\left( {\frac{1}{4} + {\frac{2}{3}}} \right)} = {\frac{{11kq}}{{12}}} = {\frac{{11q}}{{48{\pi}\varepsilon_{0}L}}}}}$
6. Two charges $\displaystyle {+q}$ each are kept $\displaystyle {2a}$ distance apart. A third charge $\displaystyle {-2q}$ is placed midway between them. Potential energy of system is:
Explanation: $\displaystyle {U = {\frac{{1}}{{4\pi\varepsilon_0}}} {\left[ {\frac{{q^2}}{{2a}}} + {\frac{{q(-2q)}}{{a}}} + {\frac{{q(-2q)}}{{a}}} \right]} = {\frac{{1}}{{4\pi\varepsilon_0}}} {\left[ {\frac{{q^2}}{{2a}}} - {\frac{{4q^2}}{{a}}} \right]} = {\frac{{-7 q^2}}{{8\pi\varepsilon_0 a}}}}$.
7. An electron accelerated through potential difference $200\text{ V}$ acquires velocity $8.4 \times 10^6\text{ m/s}$. Specific charge $e/m$ of electron is:
Explanation: $\displaystyle {e V = {\frac{{1}}{{2}}} m v^2 \implies {\frac{{e}}{{m}}} = {\frac{{v^2}}{{2V}}} = {\frac{{(8.4 \times 10^6)^2}}{{400}}} = 1.76 \times 10^{11}\text{ C/kg}}$.
8. The work done in moving a charge of $\displaystyle {4\text{ C}}$ from potential $\displaystyle {10\text{ V}}$ to $\displaystyle {30\text{ V}}$ is:
Explanation: $\displaystyle {W = q \Delta V = 4 \times (30 - 10) = 80\text{ J}}$.
9. Parallel plate capacitor plate area $\displaystyle {4 \times 10^{-2}\text{ m}^2}$, separation $\displaystyle {2\text{ mm}}$. Capacitance in air is:
Explanation: $\displaystyle {C = {\frac{{\varepsilon_0 A}}{{d}}} = {\frac{{8.854 \times 10^{-12} \times 4 \times 10^{-2}}}{{2 \times 10^{-3}}}} = 1.77 \times 10^{-11}\text{ F}}$.
10. A capacitor of $\displaystyle {4\,\mu\text{F}}$ is charged to $\displaystyle {100\text{ V}}$. Energy stored in it is:
Explanation: $\displaystyle {U = {\frac{{1}}{{2}}} C V^2 = {\frac{{1}}{{2}}} \times 4 \times 10^{-6} \times (100)^2 = 0.02\text{ J}}$.
ASSIGNMENT – 3 (ASSERTION & REASON)
1. Assertion (A): Electric potential of earth is taken as zero.
Reason (R): No electric field exists on earth surface.
Explanation: Earth potential is taken as reference zero due to huge capacitance, though electric field exists on earth.
2. Assertion (A): Work done in moving a charge between two points in uniform electric field is independent of path.
Reason (R): Electrostatic forces are not conservative.
Explanation: Path independence is true precisely because electrostatic forces ARE conservative.
3. Assertion (A): Metallic shield in form of hollow shell blocks electric field.
Reason (R): Electric field inside hollow conductor is zero at every point.
4. Assertion (A): Introducing a dielectric into isolated charged capacitor decreases energy stored.
Reason (R): Voltage decreases while charge remains constant, reducing $\displaystyle {U = Q^2 / 2C}$.
5. Assertion (A): Capacitance of parallel plate capacitor increases when dielectric slab inserted.
Reason (R): Dielectric reduces effective electric field between plates.
ASSIGNMENT – 4 (CASE STUDY QUESTIONS)
Case Study 1: Faraday Cage & Electrostatic Shielding
A Faraday cage is an enclosure made of conducting material. Fields within conductor cancel out external fields.
Q1. Best material for Faraday cage: (a) Plastic (b) Glass (c) Copper (d) Wood
Q2. Application of electrostatic shielding: (a) Microwave oven (b) Electric motor (c) Transformer
Q3. Net electric field inside hollow spherical conductor: (a) Maximum (b) Zero (c) Infinite
Case Study 2: Role of Dielectrics in Capacitors
Dielectrics increase energy storage capability by reducing internal electric field.
Q1. Polar molecule example: (a) $\displaystyle {\text{O}_2}$ (b) $\displaystyle {\text{H}_2}$ (c) $\displaystyle {\text{N}_2}$ (d) $\displaystyle {\text{HCl}}$
Q2. Correct statement about dielectrics: (a) Free electrons flow (b) Induced dipole moment aligns along applied field
Q3. When dielectric inserted into connected capacitor, energy stored: (a) Decreases (b) Increases K times
ASSIGNMENT – 5 (CONCEPTUAL & DERIVATIONS)
SHORT ANSWER QUESTIONS
1. Electric potential is constant in a given region. What can you say about electric field there?
2. Does a proton move from lower to higher or higher to lower potential region in electric field?
3. Why can two equipotential surfaces never intersect?
4. Three points A, B and C lie in uniform electric field $\displaystyle {E = 5 \times 10^3\text{ N/C}}$ as shown. Find potential difference between (i) A and B, (ii) A and C.
(ii) BC is perpendicular to field lines ($\displaystyle {V_B = V_C}$). Thus $\displaystyle {V_A - V_C = V_A - V_B = 200\text{ V}}$.
LONG ANSWER QUESTIONS / DERIVATIONS
1. Derive expression for electrostatic potential due to a point charge. Draw graph of $V$ vs $r$.
$\displaystyle {V = -\int_{\infty}^{r} E \, dr = -\int_{\infty}^{r} {\frac{{1}}{{4\pi\varepsilon_0}}} {\frac{{q}}{{r^2}}} dr = {\frac{{1}}{{4\pi\varepsilon_0}}} {\frac{{q}}{{r}}}}$. Graph of $V$ vs $r$ is rectangular hyperbola.
2. Derive expression for capacitance of parallel plate capacitor with dielectric medium K. Factors on which capacitance depends?
Depends on: (i) Area of plates $A$, (ii) Separation distance $d$, (iii) Dielectric constant $K$.
3. Two identical capacitors $C_1, C_2$ of $1\,\mu\text{F}$ connected to $6\text{ V}$ battery. Switch S closed, then opened, dielectric $K=3$ inserted. Charge and potential difference on $C_1, C_2$?
When S opened: $C_1$ remains connected to $6\text{ V}$ battery $\implies \displaystyle {V_1' = 6\text{ V}, q_1' = C_1' V_1 = (3\,\mu\text{F}) \times 6\text{ V} = 18\,\mu\text{C}}$.
$C_2$ isolated $\implies \displaystyle {q_2' = 6\,\mu\text{C}}$ (constant),$\displaystyle {{V_{2}' = {\frac{{q_{2}'}}{{C_{2}'}}} = {\frac{{6\,\mu\text{C}}}{{3\,\mu\text{F}}}} = 2\,\text{V}}}$.
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