Electrostatics Physics Electrostatic Potential and Capacitance Physics Class 12th

A 5µF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5µF capacitor. If the energy change during the charge redistribution is X/100 J then value of X to the nearest integer is:

6 Views | Posted a month ago

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Heat loss; ΔH = U? - U? = 1/2 (C? / (C? +C? ) (V? -V? )²
= 1/2 × (5×2.5)/ (5+2.5) (220-0)²µJ
= 5/ (2×3) × 22×22×100×10? J
= 5×11×22/3 × 10? J = 1210/3 × 10? J = 1210/3 × 10? ³ J × 4 × 10? ²
According to questions
x/100 = 4×10? ²
So, x=4
Note: But given answer by JEE Main x=36

0 Upvotes 0 Downvotes

Similar Questions for you

A parallel plate capacitor is charged and then charging battery is disconnected. If the plates are now pulled apart with insulated handles

alok kumar singh

Charge remains same

Energy stored in circuit 1 is E. If capacitors in circuit 1 and circuit 2 are connected in parallel as shown, the energy stored becomes $\frac{x E}{6}$ , find x.

alok kumar singh

Charge on C₁ = CV

Charge on C₂ = 4CV

When connected in parallel

$V_{C} = \frac{5 V}{3}$

$Q'_{1} = \frac{5}{3} C V$ $Q'_{2} = \frac{10}{3} C V$

$?$ $E = \frac{1}{2} C V^{2}$

$E' = \frac{25}{18} C V^{2} + \frac{25}{9} C V^{2}$

$\frac{25}{6} C V^{2} = \frac{50 E}{6}$

x = 50

Potential due to electric dipole on axial position at distance r from dipole is proportional to (assume r >> length of dipole)

alok kumar singh

$| E | = \frac{2 k P}{r^{3}}$

$E = - \frac{d v}{d r}$ , $v \propto \frac{1}{r^{2}}$

The force between two charged particle placed in air at separation x is F₀. Both the charged particle immerged in a medium of dielectric constant K without changing separation between two charge, then net force on one of the particle is now

alok kumar singh

In air $F = \frac{1}{4 π \in_{0}} \frac{q_{1} q_{2}}{r^{2}}$

In medium $F' = \frac{1}{4 π (k \in_{0})} \frac{q_{1} q_{2}}{r^{2}}$

$F' = \frac{F_{0}}{K}$

Five metallic plates each of area A and separated from adjacent plate by a distance d as shown in figure. The capacitance of the system across A and B is: