Electrostatic Potential and Capacitance

Electrostatic Potential and CapacitanceNEET Physics · Class 12 · NCERT Chapter 2

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7 interactive concept widgets for Electrostatic Potential and Capacitance. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

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Potential of a point charge

Potential due to multiple charges

E vs V for a point charge

Parallel plate capacitor

Capacitors in series and parallel

Effect of dielectric (battery vs isolated)

Energy stored in a capacitor

Potential and field

Potential of a point charge, scalar superposition for many charges, and the link between E and V.

Potential

Potential of a point charge

V = k q over r. Slide q and r to see how potential scales.

Potential of a point charge falls as 1/r. Sign matches the sign of the charge. V is a scalar (not a vector).

Charge q: 5.00 μC

Distance r: 50.0 cm

Potential V at distance r

9.000e+4 V

V = \frac{k q}{r}

Try this

Doubling r halves V (1/r scaling, slower than field's 1/r²).
Negative q gives negative V (work needed to bring a positive charge from infinity to a region of attraction is negative).
V at infinity is taken as zero by convention.

Potential superposition

Potential due to multiple charges

Add scalar contributions, with sign. Much simpler than the field calculation.

Potential due to multiple charges is the algebraic (signed) sum, not a vector sum. Add up the contributions.

q₁: 3.00 μC, r₁: 0.30 m

q₂: -3.00 μC, r₂: 0.50 m

Total V

3.600e+4 V

V = V_{1} + V_{2} = \frac{k q _{1}}{r _{1}} + \frac{k q _{2}}{r _{2}}

V₁

9.00e+4 V

V₂

-5.40e+4 V

Try this

Equal and opposite charges at equal distance: V = 0 (cancels). On the equator of a dipole, V is always 0.
Two equal positive charges: contributions add up.
V is a scalar: just add with sign. No vector tricks needed.

E and V

E vs V for a point charge

Side by side: V scales as 1/r, E as 1/r². E is the negative slope of V.

For a point charge: V falls as 1/r, E falls as 1/r². Their relationship is E = -dV/dr (E is the negative slope of V).

Charge q: 1.00 μC

● V (1/r)

● E (1/r²) dashed

E = - \frac{d V}{d r}

Try this

V curve is steeper at small r, which is where E (its negative slope) is large.
V → 0 at infinity. E → 0 at infinity (but faster).
For negative q: both V and E flip sign.

Capacitor basics

Parallel-plate capacitor and how it changes when capacitors are combined or filled with a dielectric.

Parallel plate

Parallel plate capacitor

C = K epsilon_0 A over d. Three knobs: plate area, separation and dielectric constant.

Capacitance of a parallel-plate capacitor. Larger area: more capacitance. Bigger separation: less. Higher dielectric constant K: K times more.

Plate area A: 100 cm²

Separation d: 1.00 mm

Dielectric constant K: 1.00 (1 = vacuum, 80 = water)

Capacitance

88.54 pF

C = \frac{K ϵ _{0} A}{d}

Try this

Halve d at fixed A: capacitance doubles.
Real capacitors use thin dielectrics (K of 4 to 80) and large rolled-up area.
Water has K around 80 but isn't practical (it conducts when impure).
Even with K = 1 and area 1 cm², parallel plate capacitance is tiny (a few pF).

Combinations

Capacitors in series and parallel

The combination rules for capacitors are the OPPOSITE of resistors. Memory hook: capacitors and resistors swap rules.

Three capacitors combined two ways. Series gives a smaller effective C; parallel gives a larger one.

C₁: 2 μF

C₂: 4 μF

C₃: 6 μF

Effective capacitance

1.091 μF

\frac{1}{C _{eff}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}

Try this

Series: same charge on each, voltages add. Like resistors in parallel.
Parallel: same voltage across each, charges add. Like resistors in series.
In series, C_eff is smaller than the smallest C. In parallel, larger than the largest C.

Dielectric

Effect of inserting a dielectric

Same capacitor, two scenarios. Battery still connected vs disconnected (isolated). The outcome is opposite for V and U.

Inserting a dielectric of constant K multiplies capacitance by K. What changes next depends on whether the battery is still connected.

K: 4.00 (1 = vacuum)

C₀ (no dielectric): 10 pF

V₀: 100 V

Before

C = 10.0 pF

Q = 1.00 nC

V = 100 V

U = 50.00 nJ

After

C = 40.0 pF

Q = 4.00 nC

V = 100.00 V

U = 200.00 nJ

V_{const} : Q \to K Q_{0}, U \to K U_{0}

Try this

Battery connected: V is fixed. C goes up by K, so Q and U also rise by factor K.
Isolated: Q is fixed. C goes up by K, so V and U DROP by factor K.
Energy U is conserved only if the system is isolated AND no dielectric is moved (work is done!).

Energy stored

Three equivalent forms of capacitor energy.

Energy

Energy stored in a capacitor

Three equivalent forms of the same energy: ½ Q V, ½ C V², ½ Q² over C. Pick whichever the problem hands you.

Energy in a capacitor lives in the electric field between the plates. Three equivalent formulas, pick whichever is most convenient.

C: 10.0 μF

V: 50 V

Charge stored Q

0.500 mC

Energy U

0.013 J

½ Q V

0.013

½ C V²

0.013

½ Q²/C

0.013

U = \frac{1}{2} Q V = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C}

Try this

Doubling V at fixed C: energy quadruples (V²).
Doubling C at fixed V: energy doubles (linear in C).
A 100 µF cap at 100 V holds 0.5 J, small but enough to deliver a noticeable shock.
Energy density between plates: u = ½ ε₀ E².

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