In the given figure, the charge stored in $6mutextF$ $6\mu\text{F}$ capacitor, when points A and B are joined by a connecting wire is ________ $mutextC$ $\mu\text{C}$.
The diagram displays a bridge-like network containing a 6 Ohm resistor, 3uF capacitor, 6uF capacitor, and 3 Ohm resistor powered by a 9V supply.

Numerical Answer Type:

Enter a numerical value Answer: 36 to 36 +4 marks

Solution & Explanation

### Related Formula At steady state, a capacitor acts as an open circuit to DC current. The charge on a capacitor is:

Q = C Delta V

### Core Logic When node

A

and node

B

are connected by a wire, they reach the same electrical potential (

V_A = V_B = V_M

). At DC steady state, capacitors block current, so current only flows through the resistors from the

9text V

source to ground: * The

6\ Omega

resistor is connected between

9text V

and node

M

. * The

3\ Omega

resistor is connected between node

M

and ground. Hence, the equivalent series resistance for the DC current path is:

R_texteq = 6 + 3 = 9\ Omega

The steady state current is:

I = fracVR_texteq = frac9text V9\ Omega = 1text A

Simplified equivalent steady-state circuit for Q56 — The diagram displays a bridge-like network containing a 6 Ohm resistor, 3uF capacitor, 6uF capacitor, and 3 Ohm resistor powered by a 9V supply.

### Step 1: Calculate the Potential Difference and Charge The potential at node

M

(

V_M

) is:

V_M = I times 3\ Omega = 1text A times 3\ Omega = 3text V

The

6mutextF

capacitor is connected between the

9text V

source and node

B

(which is at potential

V_M = 3text V

). Thus, the potential difference across the

6mutextF

capacitor is:

Delta V_6mutextF = 9text V - 3text V = 6text V

The charge stored in the

6mutextF

capacitor is:

Q = C Delta V = 6mutextF times 6text V = 36\ mutextC

Thus, the charge stored is

36\ mutextC

. ### Pattern Recognition Shorting

A

and

B

makes the network a simple voltage divider for resistors at steady state. Once potential of the middle node is found (

3text V

), the capacitor charge is calculated instantly using

Q = C Delta V

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatic Potential and Capacitance

Reference Study Guides

Electrostatic Potential And Capacitance

More Electrostatic Potential and Capacitance Previous-Year Questions — Page 2

Q4 jee_main_2025_28_jan_morning Equipotential Surfaces

Three infinitely long wires with linear charge density

lambda

are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?

A. $mathrmxy + mathrmyz + mathrmzx = textconstant$
B. $(mathrmx + y)(mathrmy + z)(mathrmz + x) = textconstant$
C. $(mathrmx^2 +mathrmy^2)(mathrmy^2 +mathrmz^2)(mathrmz^2 +mathrmx^2) = textconstant$
D. $mathrmxyz = textconstant$ $\mathrm{xyz} = \text{constant}$

Solution

### Related Formula

mathrmV = -int vecmathrmEcdotmathrmdvecmathrmr = 2mathrmklambda ln mathrmr + mathrmc

### Core Logic The potential at a point due to a line charge on an axis is proportional to the logarithm of its perpendicular distance. For the wire along the z-axis:

mathrmV_z = -mathrmklambda ln(x^2 + y^2)

For the wire along the x-axis:

mathrmV_x = -mathrmklambda ln(y^2 + z^2)

For the wire along the y-axis:

mathrmV_y = -mathrmklambda ln(z^2 + x^2)

Summing the individual potentials to find the net configuration potential:

mathrmV_textnet = -mathrmklambda left[ ln(x^2+y^2) + ln(y^2+z^2) + ln(z^2+x^2) right] + mathrmC'

mathrmV_textnet = -mathrmklambda ln left[ (x^2+y^2)(y^2+z^2)(z^2+x^2) right] + mathrmC'

For an equipotential surface, set

mathrmV_textnet = textconstant

: ### Step 1: Final Expression

(x^2 + y^2)(y^2 + z^2)(z^2 + x^2) = textconstant

This maps perfectly to option (3). ### Pattern Recognition Logarithmic combination rules transform scalar potential additions into products inside the functional argument:

sum ln(mathrmr_i^2) = ln(prod mathrmr_i^2)

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatic Potential and Capacitance

Q7 jee_main_2025_24_jan_evening Electrostatic Potential Energy

In the first configuration (1) as shown in the figure

Electrostatic potential energy configuration 1 square charges Q7 — The diagram displays configuration 1 with 4 charges on corners and configuration 2 with charges on the midpoints of the square sides.

, four identical charges

(q_0)

are kept at the corners A, B, C and D of square of side length 'a'. In the second configuration (2)

, the same charges are shifted to mid points G, E, H and F, of the square. If

K=frac14pivarepsilon_0

, the difference between the potential energies of configuration (2) and (1) is given by:

A. $fracKq_0^2aleft(4sqrt2-2 ight)$
B. $fracKq_0^2aleft(3-sqrt2 ight)$
C. $fracKq_0^2aleft(4-2sqrt2 ight)$
D. $fracKq_0^2aleft(3sqrt2-2 ight)$

Solution

### Related Formula

U = sum_i < j fracK q_i q_jr_ij

### Core Logic For configuration (1) with side length

a

: - 4 pairs of adjacent side-charges with distance

a

. - 2 pairs of diagonal charges with distance

sqrt2a

U_1 = 4 cdot fracKq_0^2a + 2 cdot fracKq_0^2sqrt2a = fracKq_0^2a(4 + sqrt2)

For configuration (2), charges lie on midpoints forming an inner square of side length

a' = fracasqrt2

: - 4 pairs at distance

fracasqrt2

. - 2 pairs at diagonal distance

a

U_2 = 4 cdot fracKq_0^2fracasqrt2 + 2 cdot fracKq_0^2a = fracKq_0^2a(4sqrt2 + 2)

### Step 1: Finding the difference

U_2 - U_1 = fracKq_0^2aleft[(4sqrt2 + 2) - (4 + sqrt2) ight] = fracKq_0^2a(3sqrt2 - 2)

### Pattern Recognition The side of the new square formed by midpoints is scaled down by

1/sqrt2

. Therefore, interaction terms scale accordingly based on system geometric dimensions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatic Potential and Capacitance

Q35 jee_main_2024_01_february_morning Capacitors and Capacitance

Two identical capacitors have same capacitance

C

. One of them is charged to the potential

V

and other to the potential

2V

. The negative ends of both are connected together. When the positive ends are also joined together, the decrease in energy of the combined system is:

A. $frac14 CV^2$ $\frac{1}{4} CV^2$
B. $2 CV^2$ $2 CV^2$
C. $frac12 CV^2$ $\frac{1}{2} CV^2$
D. $frac34 CV^2$ $\frac{3}{4} CV^2$

Solution

### Related Formula Energy stored in a capacitor:

U = frac12CV^2

Common potential after sharing charges:

V_C = fracq_1 + q_2C_1 + C_2

Loss of energy:

Delta U = frac12 fracC_1 C_2C_1 + C_2 (V_1 - V_2)^2

### Core Logic Initial energy of the individual systems:

U_i = frac12CV^2 + frac12C(2V)^2 = frac12CV^2 + 2CV^2 = frac52CV^2

Common potential

V_C

when joined in parallel:

V_C = fracCV + C(2V)C + C = frac3CV2C = frac32V

### Step 1: Calculate Final Energy and Loss Final energy of the combined system:

U_f = frac12(2C)V_C^2 = C left(frac32Vright)^2 = frac94CV^2

Decrease in energy (loss):

Delta U = U_i - U_f = frac52CV^2 - frac94CV^2 = frac14CV^2

### Pattern Recognition Direct Formula Shortcut:

Delta U = frac12 fracC cdot C2C (2V - V)^2 = frac14C(V)^2 = frac14CV^2

### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatic Potential and Capacitance

Q57 jee_main_2024_30_jan_morning Energy Loss in Capacitors

A capacitor of capacitance C and potential V has energy E. It is connected to another capacitor of capacitance 2C and potential 2V. Then the loss of energy is

fracx3E

, where x is ________.

Numerical Answer. Answer: 2 to 2

Solution

### Related Formula

E = frac12 C V^2

Delta E = frac12 fracC_1 C_2C_1 + C_2 (V_1 - V_2)^2

### Core Logic When two charged capacitors are connected in parallel, charge redistributes until they reach a common potential. During this redistribution, energy is dissipated as heat, strictly governed by the standard loss formula. ### Step 1: Assign Values Initial energy of first capacitor:

E = frac12 C V^2

Capacitor 1:

C_1 = C

V_1 = V

Capacitor 2:

C_2 = 2C

V_2 = 2V

### Step 2: Calculate Energy Loss

Delta E = frac12 frac(C)(2C)C + 2C (V - 2V)^2

Delta E = frac12 frac2C^23C (-V)^2

Delta E = frac12 left(frac2C3right) V^2

Delta E = frac23 left(frac12 C V^2right)

Delta E = frac23 E

### Step 3: Match the Pattern Given loss is

fracx3E

. Comparing, we get

x = 2

. ### Pattern Recognition The loss formula

Delta E = frac12 C_texteq(Delta V)^2

elegantly bypasses recalculating common potential

V_c

and summing final state energies. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatic Potential and Capacitance

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