Every problem broken into the same six moves — what you're Given, what's Asked, the Concept behind it, the Method, an Easy Trick for speed, and the Solution.
QuestionThree charges \(2q,\,-q,\,-q\) are located at the vertices of an equilateral triangle. At the centre of the triangle, the field and potential are — (a) field zero, potential non-zero; (b) field non-zero, potential zero; (c) both zero; (d) both non-zero.
centre is equidistant from all three vertices
Given
Charges \(2q,\,-q,\,-q\) at the vertices of an equilateral triangle.
Asked
Nature of field and potential at the centre.
Concept
Potential is a scalar (algebraic sum). Field is a vector (cancels only with symmetry).
Method
Centre is equidistant (distance \(r\)).
\[ V=\frac{k}{r}(2q-q-q)=0 \]
Charges aren't symmetric, so resolving the three field vectors leaves a net:
\[ E_{\text{net}}=\frac{3kq}{r^2}\neq 0 \]
Trick
If the charges add to zero → potential is zero. If they're not symmetric → field survives. "Sum zero but lopsided" = V 0, E non-zero.
Solution
AnswerField non-zero, potential zero
Q2Assertion–Reason
Question(A): The strength of the electric field in a charged and isolated capacitor is decreased when a dielectric slab is inserted. (R): When a dielectric slab is inserted between the plates of a charged capacitor, an electric field is generated due to induced charge, opposite to the external field. Choose the correct option.
Given
A: field in a charged, isolated capacitor drops when a dielectric is inserted. R: induced charge makes a field opposite to the external one.
Asked
Truth of A, R and whether R explains A.
Concept
Isolated ⇒ charge \(Q\) fixed; dielectric gives \(E=E_0/K\). Polarisation is the cause.
Method
\(E=E_0/K \lt E_0\) → A true. The induced (bound) charges set up the opposing field that reduces the net field → R true and it is the reason.
Trick
"Isolated = \(Q\) constant; dielectric always weakens the inside field." The induced charge is *why*.
Solution
AnswerBoth true; R correctly explains A
Q3Capacitance of a body
QuestionA body is charged to a certain potential. When an additional charge of \(60\text{ nC}\) is imparted to it, the rise in potential is found to be \(15\text{ mV}\). Find the capacitance of the body.
nC ÷ mV gives µF straight away (\(10^{-9}/10^{-3}=10^{-6}\)). Just do \(60/15=4\).
Solution
Answer\(4\ \mu\text{F}\)
Q4Force between plates
QuestionA parallel plate capacitor with plate area \(100\text{ cm}^2\) and separation \(1.0\text{ cm}\) is connected across a battery of emf \(24\text{ V}\). The force of attraction between the plates is —
Force \(=\tfrac12\varepsilon_0E^2A\) with \(E=V/d\). The "half" is the giveaway — never forget the \(2\) in the denominator.
Solution
Answer\(2.5\times10^{-7}\ \text{N}\)
Q5Charge sharing · spheres
QuestionTwo metal spheres having diameters in the ratio \(2:3\) are put into contact and a charge of \(20\ \mu\text{C}\) is given to the system. They are then separated so there is no mutual force between them. The potential due to the larger sphere at a distance of \(6\text{ m}\) outside the sphere is —
Given
Diameter ratio \(2:3\); touched; total \(20\ \mu\text{C}\); distance \(6\text{ m}\).
Asked
Potential due to the larger sphere at \(6\text{ m}\).
Concept
In contact → same potential → \(Q\propto R\) (since \(V=kQ/R\)).
Charge divides in the ratio of radii (= ratio of diameters). Bigger sphere = bigger share.
Solution
Answer\(18\ \text{kV}\)
Q6Dielectric in equilibrium
QuestionThe width of each plate is \(b\). The capacitor plates are rigidly clamped and connected to a battery of emf \(V\); all surfaces are frictionless. A dielectric slab is linked over a pulley to a hanging mass \(M\). Find the value of \(M\) for which the dielectric slab stays in equilibrium.
slab pulled in by the field, balanced by weight Mg
Given
Width \(b\), constant \(K\), gap \(d\), emf \(V\); slab linked over a pulley to mass \(M\).
Asked
\(M\) for which the slab stays in equilibrium.
Concept
At constant \(V\), the capacitor pulls the slab in with constant force \(F=\dfrac{\varepsilon_0 bV^2(K-1)}{2d}\).
Constant-voltage slab force \(\propto (K-1)\). Equate to weight \(Mg\) and solve.
Solution
Answer\(M=\dfrac{\varepsilon_0 bV^2}{2dg}(K-1)\)
Q7Energy loss · infinite chain
QuestionAll spheres are identical and initially uncharged. \(S_1\) is closed at \(t=0\), leaving the others open. After a long time \(S_1\) is opened and \(S_2\) is closed; the process repeats until \(S_{n-1}\) is opened and \(S_n\) is closed. Find the net loss of energy of the system if it is made of an infinite number of spheres. (\(R\) = radius of each sphere.)
charge handed from one sphere to the next, one switch at a time
Given
Identical spheres, \(C=4\pi\varepsilon_0 R\); sphere 1 charged by \(V\), then charge passed down the chain; infinite spheres.
Asked
Total energy lost.
Concept
Two losses: (1) charging from the battery wastes half; (2) each equal-sphere share wastes half of what remains.
Method
Charging loss \(=\tfrac12CV^2\). Sharing losses form a GP:
QuestionConsider a point charge \(q\). The equipotential surfaces whose potentials differ by a constant amount (say \(1.0\text{ volt}\)) are — equally spaced / unequally spaced with separation increasing away from \(q\) / unequally spaced with separation decreasing away from \(q\) / none of these.
Given
Point charge \(q\); equipotential surfaces differing by \(1\text{ V}\).
Asked
Are they equally or unequally spaced?
Concept
Surfaces are spheres \(r=\dfrac{kq}{V}\); \(V\propto\dfrac1r\).
Method
Equal voltage steps need bigger \(\Delta r\) as you move out, because the field weakens.
Trick
Weaker field far away → more distance per volt → spacing grows outward.
Solution
AnswerUnequal — spacing increases away from \(q\)
Q9Dipole · non-uniform field
QuestionA dipole (\(+q\) on the left, \(-q\) on the right) is placed in a non-uniform electric field whose lines fan outward toward the right. In which direction will it move, and what happens to its potential energy?
field lines spread out to the right → field is stronger on the left
Given
Dipole \(+q\) (left), \(-q\) (right); field lines fan outward to the right → field stronger on the left.
Asked
Direction of motion and change in PE.
Concept
Net force = bigger force wins; \(U=-\vec p\cdot\vec E\).
Method
\(+q\) sits in the strong field → large force to the right.
\(-q\) sits in the weak field → small force to the left.
Net force → right. Here \(\vec p\) (from \(-q\) to \(+q\)) is opposite \(\vec E\), so \(U=+pE\); moving into weaker \(E\) makes \(U\) decrease.
Trick
The charge sitting in the stronger field wins — the dipole drifts that way, and a freely moving dipole always loses PE as it moves.
Solution
AnswerMoves right; PE decreases
Q10Zero interaction energy
QuestionThree equal charges \(Q\) are placed at the three vertices of an equilateral triangle. What should be the value of a charge that, when placed at the centroid, reduces the interaction energy of the system to zero?
find q at the centroid so total energy = 0
Given
Three equal \(Q\) at vertices of an equilateral triangle (side \(a\)); charge \(q\) at the centroid.
Asked
\(q\) that makes total interaction energy zero.
Concept
\(U_{\text{total}}=U_{\text{(three }Q)}+U_{\text{(}q\text{ with each }Q)}=0\).
Centroid-to-vertex \(=\dfrac{a}{\sqrt3}\) — that \(\sqrt3\) is exactly what lands in the answer.
Solution
Answer\(q=-\dfrac{Q}{\sqrt3}\)
Q11Switch in a circuit
QuestionCalculate the charge on the second capacitor before and after the switch in the circuit is closed. (Two equal capacitors \(C\), battery emf \(E\); closing the switch shorts the first capacitor.)
closing the switch shorts ① → full emf falls across ②
Given
Two equal capacitors \(C\), emf \(E\); the switch shorts the first capacitor when closed.
Asked
Charge on the second capacitor, before and after.
Concept
Series before closing; full emf on the survivor after the short.
Method
Before: series equal caps → \(q=\left(\tfrac C2\right)E=\dfrac{CE}{2}\).
After: cap 1 shorted → full \(E\) across cap 2 → \(q=CE\).
Trick
Equal caps in series split voltage equally (→ \(CE/2\)); shorting one dumps the whole emf on the other (→ \(CE\)).
Solution
Answer\(\dfrac{CE}{2}\) before, \(CE\) after
Q12Force on dielectric · isolated
QuestionTwo square plates (side \(\ell\), gap \(d\)) are charged to potential difference \(V_0\) and isolated. A dielectric of permittivity \(\varepsilon_r=2\), thickness \(d\) and width equal to the plates is drawn into the gap (its length \(>\ell\)). The force on the dielectric at \(x=\ell/2\) is \(\dfrac{\alpha CV_0^2}{\beta\ell}\), where \(C=\dfrac{\varepsilon_0\ell^2}{d}\). With \(\alpha,\beta\) as minimum integers, find \(\beta-\alpha\).
overlap x = ℓ/2; force pulls the slab further in
Given
Square plates (side \(\ell\), gap \(d\)) charged to \(V_0\), isolated; slab \(\varepsilon_r=2\); evaluate at \(x=\ell/2\). \(F=\dfrac{\alpha CV_0^2}{\beta\ell}\), \(C=\dfrac{\varepsilon_0\ell^2}{d}\).
Asked
\(\beta-\alpha\) (smallest integers).
Concept
Overlap → two parallel capacitors. Isolated ⇒ fixed \(Q\): \(F=\dfrac{Q^2}{2C^2}\dfrac{dC}{dx}\).
\(\dfrac{dC}{dx}\) is constant; only \(C(x)^2\) in the denominator changes with position. Plug \(x=\ell/2\) once.
Solution
Answer\(\beta-\alpha=7\)
Q13Two statements
QuestionStatement I: The value of electric potential and electric intensity at the middle point of the line joining an electron and a proton is zero. Statement II: Electric potential is a vector and electric intensity is a scalar quantity. Which is/are correct?
Given
I: at the midpoint between an electron and a proton, both \(V\) and \(E\) are zero. II: potential is a vector, field is a scalar.
Asked
Which statements are correct.
Concept
\(V\) (scalar) can cancel; \(E\) (vector) between unlike charges adds; II swaps the definitions.
Method
Midpoint: \(V=\dfrac{k(+e)}{r}+\dfrac{k(-e)}{r}=0\), but both field vectors point toward the electron → \(E\neq0\). So I is wrong. II is wrong (potential = scalar, field = vector).
Trick
Between unlike charges the two fields point the same way and add. And "potential = vector" is always a planted error.
Solution
AnswerBoth statements false
Q14Assertion–Reason
QuestionAssertion: For practical purposes, the earth is used as a reference at zero potential in electrical circuits. Reason: The electrical potential of a sphere of radius \(R\) with charge \(Q\) uniformly distributed on its surface is \(\dfrac{Q}{4\pi\varepsilon_0 R}\). Choose the correct option.
Given
A: earth is the zero-potential reference. R: \(V=\dfrac{Q}{4\pi\varepsilon_0R}\) for a charged sphere.
Asked
Truth of A, R and the link.
Concept
Earth ≈ 0 because it's a vast conductor / charge reservoir, not merely because of the sphere formula.
Method
Both statements are individually correct, but the sphere-potential formula doesn't justify the convention of choosing earth as zero — so R isn't the explanation.
Trick
A "true reason" that doesn't actually cause the assertion → "both true, not the explanation."
Solution
AnswerBoth true; R is not the correct explanation
Q15Series stack · shrinking area
QuestionConducting plates are placed face to face with consecutive separation \(d\). Plate areas are \(A,\ \tfrac A2,\ \tfrac A4,\ \dots,\ \tfrac{A}{2^{\,n-1}}\). A dielectric slab of constant \(k\) is included and the assembly is charged by a battery of emf \(\varepsilon\). Find the charge in the assembly.
Given
Plate areas \(A,\tfrac A2,\dots,\tfrac{A}{2^{n-1}}\); gap \(d\); one slab \(k\); emf \(\varepsilon\).
Asked
Charge on the assembly.
Concept
Consecutive plates → capacitors in series → add reciprocals; one carries the dielectric \(k\).
QuestionAn electron is released from a distance of \(4\text{ m}\) from a stationary point charge \(20\text{ nC}\). What is its speed when it is \(2\text{ m}\) from the charge? (\(e=1.6\times10^{-19}\text{ C}\), \(m=9\times10^{-31}\text{ kg}\), \(\tfrac{1}{4\pi\varepsilon_0}=9\times10^{9}\) SI.)
Given
Electron released from rest at \(r_1=4\text{ m}\) from \(q=20\text{ nC}\); find speed at \(r_2=2\text{ m}\).
Asked
Speed of the electron.
Concept
Loss in PE = gain in KE: \(\tfrac12mv^2=kqe\!\left(\tfrac1{r_2}-\tfrac1{r_1}\right)\).
Use \(\tfrac1{r_2}-\tfrac1{r_1}=\tfrac12-\tfrac14=\tfrac14\). Keep \(kqe\) as one number and divide.
Solution
Answer\(4\times10^6\ \text{m/s}\)
Q17PE of free charges
QuestionThree charged particles are initially in position 1. They are free to move and come to position 2 after some time. Let \(U_1\) and \(U_2\) be the electrostatic potential energies in positions 1 and 2. Then —
Given
Three charges, free to move, drift from configuration 1 to 2.
Asked
Compare \(U_1\) and \(U_2\).
Concept
A free system spontaneously moves to lower PE (PE → KE).
Method
Spontaneous motion ⇒ \(U_2 \lt U_1\).
Trick
"Free to move" = nature lowers the energy. The later configuration always has less PE.
Solution
Answer\(U_1>U_2\)
Q18Plates in series
QuestionFind the equivalent capacitance between points \(A\) and \(B\) if the capacitance between any two adjacent plates is \(C\). (\(n\) plates in a row; \(A\) at the first plate, \(B\) at the last.)
n plates → (n−1) capacitors of C in series
Given
\(n\) plates in a row; adjacent-plate capacitance \(C\); terminals at the two ends.
Asked
Equivalent capacitance between \(A\) and \(B\).
Concept
Adjacent gaps form capacitors in series.
Method
\(n\) plates → \((n-1)\) capacitors of \(C\) in series → \(C_{eq}=\dfrac{C}{n-1}\).
Trick
Plates count one more than gaps: \(n\) plates = \((n-1)\) series capacitors.
Solution
Answer\(C_{eq}=\dfrac{C}{n-1}\)
Q19Network reduction
QuestionThe resultant capacitance of the given circuit (capacitors of \(2C,\ 2C,\ C\) and \(2C,\ C,\ C\) arranged between \(P\) and \(Q\)) is —
Given
Network with capacitors \(2C,2C,C\) and \(2C,C,C\) between \(P\) and \(Q\).
Asked
Resultant capacitance.
Concept
Balanced bridge — the bridging capacitor carries no charge and drops out.
Method
With the bridge removed, the two branches each reduce to \(C\); in parallel they give \(2C\).
Caveat: the exact wiring is hard to read from the screenshot — best reading is \(2C\). Send me which plates connect to \(P\), \(Q\) and the middle node and I'll re-derive it cleanly.
Trick
Check the balance ratio first; if balanced, ignore the middle capacitor and the rest is simple series–parallel.
Solution
Answer\(2C\) (verify figure)
Q20Energy lost in sharing
QuestionA condenser of capacity \(C_1\) is charged to potential \(V_0\); the electrostatic energy stored in it is \(U_0\). It is then connected in parallel to another uncharged condenser of capacity \(C_2\). The energy dissipated in the process is —
Given
\(C_1\) charged to \(V_0\) (energy \(U_0\)) joined in parallel to uncharged \(C_2\).
Asked
Energy dissipated.
Concept
Charge conserved; loss \(=\dfrac12\dfrac{C_1C_2}{C_1+C_2}V_0^2\).
Energy lost = \(\dfrac{\text{(the other cap)}}{\text{(sum of caps)}}\times\) initial energy.
Solution
Answer\(\Delta U=\dfrac{C_2}{C_1+C_2}U_0\)
Q21Common potential
QuestionA parallel plate capacitor with air as dielectric is charged to potential \(V\) using a battery. After removing the battery, the charged capacitor is connected across an identical uncharged parallel plate capacitor filled with wax of dielectric constant \(k\). The common potential of both capacitors is —
Given
Air capacitor charged to \(V\) (battery removed); connected to an identical uncharged capacitor filled with wax \(k\).
Same plates ⇒ \(C_2=kC_1\). Common potential \(=\dfrac{V}{1+k}\).
Solution
Answer\(\dfrac{V}{k+1}\ \text{volts}\)
Q22Series of charges
QuestionAn infinite number of electric charges, each of magnitude \(2\text{ nC}\), are placed along the x-axis at \(x=1,\ 3,\ 9,\ 27\text{ cm}\dots\) and so on. If consecutive charges have opposite sign, the electric potential at \(x=0\) is —
Given
\(2\text{ nC}\) charges, alternating sign, at \(x=1,3,9,27,\dots\) cm.
Asked
Potential at the origin.
Concept
\(V=kq\sum\pm\dfrac1r\); geometric series, ratio \(-\tfrac13\).
Pull out \(\dfrac{kq}{r_1}\) and sum the GP. Alternating ratio \(\tfrac13\) → factor \(\dfrac34\).
Solution
Answer\(1350\ \text{V}\)
Q23Two series of charges
QuestionCharges of \(+5\text{C}\) are placed at \(y=1,\ 3,\ 9,\ 27\text{ cm}\dots\) and charges of \(-5\text{C}\) at \(x=1,\ 2,\ 4,\ 8\text{ cm}\dots\) (both infinite series). Find the electric potential at the origin (\(K=\tfrac{1}{4\pi\varepsilon_0}\)).
Given
\(+5\text{C}\) at \(y=1,3,9,\dots\) cm; \(-5\text{C}\) at \(x=1,2,4,\dots\) cm; \(K=\tfrac1{4\pi\varepsilon_0}\).
Asked
Potential at the origin.
Concept
Two geometric series — ratio \(\tfrac13\) (sum \(\tfrac32\)) and ratio \(\tfrac12\) (sum \(2\)).
Powers of 3 → ratio ⅓ → sum 3/2; powers of 2 → ratio ½ → sum 2. Then combine signs.
Solution
Answer\(-250K\)
Q24V from a uniform field
QuestionThe potential at the origin is zero, due to electric field \(\vec E=40\hat i+60\hat j\ \text{NC}^{-1}\). The potential at point \(P(3\text{ m},\,3\text{ m})\) is —
Given
\(V=0\) at origin; \(\vec E=40\hat i+60\hat j\ \text{N/C}\); point \(P(3,3)\).
Asked
Potential at \(P\).
Concept
For a uniform field, \(V_P=V_O-\vec E\cdot\vec r\).
Method
\[ V_P=0-(40\times3+60\times3)=-300\ \text{V} \]
Trick
From the origin, \(V=-(E_xx+E_yy)\). Mind the minus sign.
Solution
Answer\(-300\ \text{V}\)
Q25Energy density
QuestionIf \(E\) is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to —
Given
Electric field intensity \(E\) in an electrostatic field.
Asked
Energy density's dependence on \(E\).
Concept
\(u=\tfrac12\varepsilon_0E^2\).
Method
Directly, \(u\propto E^2\).
Trick
Energy density is always proportional to \(E^2\) (like \(\tfrac12CV^2\) being "square" too).