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Electrostatic Potential and Capacitance formulas

Class 12 physics formula sheet for NEET and JEE: the key equations of NCERT chapter 2, the special cases questions are built on, and diagrams where they help.

55 formulas9 sectionsClass 12 · Chapter 23 of 9 sections free

By Sreeraj P, M.Sc Physics · 10+ years teaching NEET and JEE

Most used formulasOther formulas and cases

1Electric potential

$$V=\frac Wq,\qquad V=\frac{kq}{r},\qquad V=\sum\frac{kq_i}{r_i}$$

Scalar (J C−1 = volt). Work to bring unit + charge from ∞. Potentials simply add with signs.

$$\begin{array}{l}\displaystyle \vec E=-\nabla V=-\left(\frac{\partial V}{\partial x}\hat i+\frac{\partial V}{\partial y}\hat j+\frac{\partial V}{\partial z}\hat k\right)\\[5pt]\displaystyle V_B-V_A=-\int_A^B\vec E\cdot d\vec l\end{array}$$

$\vec E$ points towards decreasing $V$. Uniform field: $E=\dfrac Vd$. Work by the field on $q$ from A to B: $q(V_A-V_B)$.

$$qV=\tfrac12mv^2,\qquad v=\sqrt{\frac{2qV}{m}}$$

Charge accelerated from rest through $V$. 1 eV $=1.6\times10^{-19}$ J.

$$\text{Zero-potential points: }\frac{q_1}{x}=\frac{q_2}{d-x},\ \ \frac{q_1}{y}=\frac{q_2}{d+y}$$
$$\begin{array}{l}\displaystyle \text{Zero-potential points: }\frac{q_1}{x}=\frac{q_2}{d-x}\\[6pt]\displaystyle \ \frac{q_1}{y}=\frac{q_2}{d+y}\end{array}$$

For $q_1$ and $-q_2$ a distance $d$ apart: one point between, one outside (beyond the smaller). Perpendicular bisector of $+q,-q$: $V=0$ everywhere.

  • Equipotential surfaces: ⟂ to field lines; no work along them; never intersect; closer together where $E$ is strong. Point charge → concentric spheres; uniform field → parallel planes.
  • Where $E=0$, $V$ need not be zero, and vice versa.

2Potential of charge distributions

$$\text{Shell (or conductor): }V=\frac{kQ}{r}\ (r\ge R),\quad V=\frac{kQ}{R}\ (r\le R)$$
$$\begin{array}{l}\displaystyle \text{Shell (or conductor): }V=\frac{kQ}{r}\ (r\ge R)\\[6pt]\displaystyle V=\frac{kQ}{R}\ (r\le R)\end{array}$$

Constant inside, equal to the surface value.

$$\text{Solid sphere: }V=\frac{kQ}{2R^3}(3R^2-r^2)\ (r
$$\begin{array}{l}\displaystyle \text{Solid sphere: }V=\frac{kQ}{2R^3}(3R^2-r^2)\ (r

Uniformly charged insulating sphere.

$$\text{Ring axis: }V=\frac{kQ}{\sqrt{R^2+x^2}},\qquad \text{centre: }\frac{kQ}{R}$$
$$\begin{array}{l}\displaystyle \text{Ring axis: }V=\frac{kQ}{\sqrt{R^2+x^2}}\\[6pt]\displaystyle \text{centre: }\frac{kQ}{R}\end{array}$$

Same even if the charge is not uniform on the ring.

$$\text{Disc axis: }V=\frac{\sigma}{2\varepsilon_0}\left(\sqrt{R^2+x^2}-x\right),\quad \text{centre: }\frac{\sigma R}{2\varepsilon_0}$$
$$\begin{array}{l}\displaystyle \text{Disc axis: }V=\frac{\sigma}{2\varepsilon_0}\left(\sqrt{R^2+x^2}-x\right)\\[6pt]\displaystyle \text{centre: }\frac{\sigma R}{2\varepsilon_0}\end{array}$$

Uniform disc of radius $R$.

$$\text{Line: }V=-\frac{\lambda}{2\pi\varepsilon_0}\ln r+C,\qquad \text{sheet: }V=-\frac{\sigma}{2\varepsilon_0}r+C$$
$$\begin{array}{l}\displaystyle \text{Line: }V=-\frac{\lambda}{2\pi\varepsilon_0}\ln r+C\\[6pt]\displaystyle \text{sheet: }V=-\frac{\sigma}{2\varepsilon_0}r+C\end{array}$$

Only differences are meaningful: $V_1-V_2=\dfrac{\lambda}{2\pi\varepsilon_0}\ln\dfrac{r_2}{r_1}$.

$$V_{\text{in}}=\frac{kq_1}{a}+\frac{kq_2}{b},\qquad V_{\text{out}}=\frac{k(q_1+q_2)}{b}$$
$$\begin{array}{l}\displaystyle V_{\text{in}}=\frac{kq_1}{a}+\frac{kq_2}{b}\\[6pt]\displaystyle V_{\text{out}}=\frac{k(q_1+q_2)}{b}\end{array}$$

Concentric shells. Inner shell earthed: its charge becomes $-q_2\dfrac ab$ (outer charge $q_2$). Outer earthed: outer surface charge becomes 0.

abq₁q₂Vinner = kq₁/a + kq₂/bVouter = k(q₁+q₂)/b
$$\text{Two spheres joined by a wire: }\frac{q_1}{q_2}=\frac{r_1}{r_2},\qquad \frac{\sigma_1}{\sigma_2}=\frac{r_2}{r_1}$$
$$\begin{array}{l}\displaystyle \text{Two spheres joined by a wire: }\frac{q_1}{q_2}=\frac{r_1}{r_2}\\[6pt]\displaystyle \frac{\sigma_1}{\sigma_2}=\frac{r_2}{r_1}\end{array}$$

Common potential $V=\dfrac{q_1+q_2}{4\pi\varepsilon_0(r_1+r_2)}$. Bubble of radius $r$, thickness $t$ at $V$ collapsing to a drop: $V'=V\left(\dfrac{r}{3t}\right)^{1/3}$.

3Dipole potential

$$V=\frac{kp\cos\theta}{r^2}=\frac{k\,\vec p\cdot\vec r}{r^3}$$

Axial: $kp/r^2$; equatorial: 0. Falls as $1/r^2$ (point charge $1/r$).

6 more sections and 32 formulas in the full chapter

  1. 4Potential energy5 formulas
  2. 5Capacitance and simple capacitors9 formulas
  3. 6Dielectric and metal slabs6 formulas · 4 diagrams
  4. 7Combination of capacitors5 formulas · 1 diagram
  5. 8Energy, force, sharing charge7 formulas
  6. 9What changes when…1 case table

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