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MHT-CET Physics · Formula sheet

Electrostatics formulas

23 formulas and 24 common traps for MHT-CET Physics Electrostatics, grouped by subtopic.

Full notes with worked examples

Coulomb's Law and Electric Field

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Coulomb's Law and Force Ratios

Coulomb's law

F=14πε0K q1q2r2F = \frac{1}{4\pi\varepsilon_0 K}\,\frac{q_1 q_2}{r^2}

Where the Field or Force Is Zero on a Line

Unlike charges: distance beyond the smaller one

∣q1∣(L+x)2=∣q2∣x2  ⇒  x=L∣q1∣/∣q2∣−1\frac{|q_1|}{(L+x)^2} = \frac{|q_2|}{x^2} \;\Rightarrow\; x = \frac{L}{\sqrt{|q_1|/|q_2|} - 1}

Superposition: Symmetric Arrangements and Field Lines

Field of a point charge

E⃗=kqr2 r^,E⃗net=∑iE⃗i\vec E = \frac{kq}{r^2}\,\hat r, \qquad \vec E_{\text{net}} = \sum_i \vec E_i

A Charge Moving in, or Held by, a Uniform Field

Motion in a uniform field

a=qEm,KE=q2E2t22m,qE=mg (balance)a = \frac{qE}{m}, \qquad \text{KE} = \frac{q^2E^2t^2}{2m}, \qquad qE = mg \text{ (balance)}

Common traps

Adding the moved charge to one side only

When charge moves from the positive body to the negative one, the positive loses it AND the negative's magnitude drops by the same amount. +4q,−4q+4q, -4q with qq moved is +3q,−3q+3q, -3q, so the force is 916F\frac{9}{16}F, not 1216F\frac{12}{16}F.

Looking between two unlike charges

Between +8q+8q and −2q-2q both fields point toward −2q-2q and can only add. The answer L4\frac{L}{4} sits in the options for students who looked there; the real point is 2L2L, beyond the smaller charge.

Zero field does not mean zero potential

At the centre of 2q,−q,−q2q, -q, -q the potential is zero because the charges sum to zero, but the fields do not cancel. The paper offers all four combinations; decide field and potential separately.

Using the electron's mass for a heavier particle

'Charge 3e3e, mass 2m2m' gives a=3eE2ma = \frac{3eE}{2m}. The options include the same expression with the charge and mass swapped; divide the particle's own charge by its own mass.

Gauss's Law and Electric Flux

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Electric Flux and Gauss's Law

Gauss's law

∮E⃗⋅dA⃗=qencε0\oint \vec E\cdot d\vec A = \frac{q_{\text{enc}}}{\varepsilon_0}

Fields of Spheres, Sheets, Cylinders and Conductors

Three results to recall

Esheet=σ2ε0,Econductor=σε0,Esolid sphere(r≤R)=ρr3ε0E_{\text{sheet}} = \frac{\sigma}{2\varepsilon_0}, \qquad E_{\text{conductor}} = \frac{\sigma}{\varepsilon_0}, \qquad E_{\text{solid sphere}}(r \le R) = \frac{\rho r}{3\varepsilon_0}

Common traps

Subtracting the wrong way round

Gauss's law counts OUTWARD flux as positive, so the net flux is leaving minus entering: q=ε0(ϕ2−ϕ1)q = \varepsilon_0(\phi_2 - \phi_1). The options pair both orders with both ε0\varepsilon_0 placements — fix the order first, then multiply.

Distance from the surface is not distance from the centre

'0.2 m from a point on the surface' of a 0.1 m sphere puts the point 0.3 m from the centre. The field formula uses the distance from the centre; the other reading lands on an option too.

Sheet versus conductor

A thin sheet of charge sends half its field each way: σ2ε0\frac{\sigma}{2\varepsilon_0}. A conductor has field only on its outside: σε0\frac{\sigma}{\varepsilon_0}. The paper offers both with both directions.

Electric Dipole — Field, Potential and Torque

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Field and Potential on the Axis and the Equator

Short dipole

Eaxis=2kpr3,Eequator=kpr3,Vaxis=kpr2E_{\text{axis}} = \frac{2kp}{r^3}, \quad E_{\text{equator}} = \frac{kp}{r^3}, \quad V_{\text{axis}} = \frac{kp}{r^2}

Torque, Energy and Work for a Dipole in a Uniform Field

Dipole in a uniform field

τ=pEsin⁡θ,U=−pEcos⁡θ,Wθ1→θ2=pE(cos⁡θ1−cos⁡θ2)\tau = pE\sin\theta, \qquad U = -pE\cos\theta, \qquad W_{\theta_1 \to \theta_2} = pE(\cos\theta_1 - \cos\theta_2)

Common traps

Field power versus potential power

Field falls as 1/r31/r^3, potential as 1/r21/r^2. A question asking 'the potential on the axis is proportional to' has 1/r31/r^3 as the planted wrong answer.

Using sin θ for the work

Torque uses sin⁡θ\sin\theta; work and energy use cos⁡θ\cos\theta. Turning an aligned dipole through 60∘60^\circ costs pE(1−cos⁡60∘)=pE2pE(1 - \cos 60^\circ) = \frac{pE}{2}, not pEsin⁡60∘pE\sin 60^\circ.

Electric Potential and Potential Energy

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Potential of Point Charges, and Equipotentials

Potential of charges

V=14πε0∑iqiriV = \frac{1}{4\pi\varepsilon_0}\sum_i \frac{q_i}{r_i}

Field From Potential, and Potential From Field

Field and potential

E=−dVdx,VB−VA=−∫ABE dxE = -\frac{dV}{dx}, \qquad V_B - V_A = -\int_A^B E\,dx

Potential Energy of a Group of Charges

Energy of a system

U=∑pairs i<jkqiqjrijU = \sum_{\text{pairs } i<j} \frac{kq_iq_j}{r_{ij}}

Speed Gained Through a Potential Difference

Energy gained

qV=12mv2,v=2qVmqV = \tfrac{1}{2}mv^2, \qquad v = \sqrt{\frac{2qV}{m}}

Merging Drops and Connected Spheres

Merging drops

Vbig=n2/3 vV_{\text{big}} = n^{2/3}\,v

Common traps

Dropping the sign of a negative charge

Potentials add with their signs. Four charges +q,+q,−q,−q+q, +q, -q, -q on a square give 2kqL(1−15)\frac{2kq}{L}\left(1 - \frac{1}{\sqrt{5}}\right) at the midpoint of the positive side; the 1+151 + \frac{1}{\sqrt{5}} option is what you get by adding magnitudes.

Forgetting the minus sign

E=−dVdxE = -\frac{dV}{dx}. Where VV rises with xx the field points along −x-x. The paper always offers the same magnitude with both signs or both directions.

Missing a pair

Three charges have THREE pairs, including the two outer charges with each other. q,−2q,qq, -2q, q on a line of length 2r2r gives −7q28πε0r-\frac{7q^2}{8\pi\varepsilon_0 r} only with the q22r\frac{q^2}{2r} end-to-end term.

Taking the charge ratio as the speed ratio

+q+q and +4q+4q of the same mass through the same V: the ENERGIES are 1 : 4, the speeds 1 : 2. The 1 : 4 option is always there.

Potential rising as n, or as the cube root of n

Charge rises nn times and radius n1/3n^{1/3} times; the potential is their ratio, n2/3n^{2/3}. Both nn and n1/3n^{1/3} are offered every time.

Capacitance and Combinations of Capacitors

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Capacitance of Plates and Spheres

Parallel-plate and spherical capacitors

C=ε0Ad,Csphere=4πε0RC = \frac{\varepsilon_0 A}{d}, \qquad C_{\text{sphere}} = 4\pi\varepsilon_0 R

Series and Parallel Combinations

Combinations

1Cs=1C1+1C2+⋯ ,Cp=C1+C2+⋯\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots, \qquad C_p = C_1 + C_2 + \cdots

Reading Capacitance From a Graph

Slope of the line

V=1C QV = \frac{1}{C}\,Q

Common traps

Half the voltage from ±V plates

Plates at +20+20 V and −20-20 V have a 40 V difference between them. Using 20 V doubles the capacitance and lands on a printed option.

Giving the bigger voltage to the bigger capacitor

In series V∝1CV \propto \frac{1}{C}: 3 μ3\,\muF and 2 μ2\,\muF across 100 V put 60 V on the 2 μ2\,\muF. Asked for V2:V1V_2 : V_1, the answer is 3:23 : 2, and 2:32 : 3 is waiting.

Steeper means bigger — on the wrong graph

With V on the y-axis, the steeper line needs more volts for the same charge: it is the SMALLER capacitor.

Dielectrics in Capacitors

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Battery Connected or Removed: What Changes

Full dielectric

C=Kε0Ad;Q fixed: U→UK;V fixed: U→KUC = \frac{K\varepsilon_0 A}{d}; \qquad Q\text{ fixed: } U \to \frac{U}{K}; \qquad V\text{ fixed: } U \to KU

A Slab Filling Part of the Gap

Partial slab

C=ε0Ad−t(1−1K)C = \frac{\varepsilon_0 A}{d - t\left(1 - \dfrac{1}{K}\right)}

Two Dielectrics: Series or Parallel From the Figure

Equal halves

stacked: C=2K1K2K1+K2C0,side by side: C=K1+K22C0\text{stacked: } C = \frac{2K_1K_2}{K_1 + K_2}C_0, \qquad \text{side by side: } C = \frac{K_1 + K_2}{2}C_0

Common traps

Not asking what is held fixed

Energy falls to UK\frac{U}{K} when the charge is fixed and rises to KUKU when the voltage is. The question tells you which only by saying 'charged and isolated' or 'battery remains connected'.

Subtracting t/K instead of t(1 − 1/K)

The slab replaces tt of air with an air-equivalent tK\frac{t}{K}, so the gap loses t−tKt - \frac{t}{K}. The options also offer t(1+1K)t\left(1 + \frac{1}{K}\right) and a factor of 2 in front.

The same words, two different figures

'Filled with two dielectrics as shown' has been set with both figures. The ratio K1+K22K1K2\frac{K_1 + K_2}{2K_1K_2} belongs to the STACKED figure; side by side gives 2K1+K2\frac{2}{K_1 + K_2}. Decide from the boundary line, not from memory of a past answer.

Energy Stored in a Capacitor

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Three Forms of the Stored Energy

Stored energy

U=12CV2=Q22C=12QV,u=12ε0E2U = \tfrac{1}{2}CV^2 = \frac{Q^2}{2C} = \tfrac{1}{2}QV, \qquad u = \tfrac{1}{2}\varepsilon_0E^2

Energy of Series and Parallel Groups

Equal energy, series and parallel

12CsVs2=12CpVp2  ⇒  VsVp=CpCs\tfrac{1}{2}C_sV_s^2 = \tfrac{1}{2}C_pV_p^2 \;\Rightarrow\; \frac{V_s}{V_p} = \sqrt{\frac{C_p}{C_s}}

Joining Two Charged Capacitors: Common Potential and Energy Lost

Common potential and loss

V=C1V1+C2V2C1+C2,ΔU=C1C22(C1+C2)(V1−V2)2V = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}, \qquad \Delta U = \frac{C_1C_2}{2(C_1 + C_2)}(V_1 - V_2)^2

Pulling the Plates Apart: Work and Force

Isolated capacitor, gap multiplied by n

W=(n−1) ε0AV22d,F=Q22ε0AW = (n - 1)\,\frac{\varepsilon_0AV^2}{2d}, \qquad F = \frac{Q^2}{2\varepsilon_0A}

Common traps

Treating energy as proportional to charge

Energy goes as Q2Q^2 (or V2V^2). A 44% rise in energy is a 20% rise in charge; the 3 C added is then 20% of the original, 15 C.

Squaring the capacitance ratio

Equal energy gives Vs2Vp2=CpCs\frac{V_s^2}{V_p^2} = \frac{C_p}{C_s}, so the VOLTAGE ratio is the square root. For four identical capacitors that is 4 : 1, while 16 : 1 is the ratio of the capacitances.

Energy conserved by assumption

The final energy is always LESS than the initial whenever the potentials differed. Computing 12(C1+C2)V2\frac{1}{2}(C_1 + C_2)V^2 and calling it the initial energy gives no loss and matches no option.

Taking the final energy as the work

The work is the INCREASE, Uf−UiU_f - U_i. Pulled to 4 times the gap, Uf=4UiU_f = 4U_i but the work is 3Ui=3ε0AV22d3U_i = \frac{3\varepsilon_0AV^2}{2d}; 2ε0AV2d\frac{2\varepsilon_0AV^2}{d} is the final energy and it is printed too.

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