Electrostatics

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New answer posted•a year ago

Electrostatics Class 12th

A two point charges $4 q$ and $- q$ are fixed on the $x$ -axis at $x = - \frac{d}{2}$ $\frac{}{}$ and $x = \frac{d}{2}$ $\frac{}{}$ , respectively. If a third point charge ' $q$ ' is taken from the origin to $x = d$ along the semicircle as shown in the figure, the energy of the charge will:

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$U_{initial} = \frac{k (4 q) (q)}{(d / 2)} + \frac{k (q) (- q)}{(d / 2)}$

$\begin{array}{r} \Rightarrow \frac{6 k q^{2}}{d} \\ \Rightarrow U_{final} = \frac{4 (4 q) (q)}{(\frac{3 d}{2})} + \frac{k (q) (- q)}{(d / 2)} \\ \Rightarrow \frac{2}{3} \frac{k q^{2}}{d} \\ \Rightarrow Δ U = (\frac{2}{3} - 6) \frac{k q^{2}}{d} \Rightarrow \frac{- 16}{3} \frac{{k q}^{2}}{d} \end{array}$

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New answer posted•a year ago

Electrostatics Class 12th

Concentric metallic hollow spheres of radii <!-- [if gte msEquation 12]>R $R$ and $4 R$ <!-- [if gte msEquation 12]>4R $Q_{1}$ and $Q_{2}$ <!-- [if gte msEquation 12]>Q2 $V (R) - V (4 R)$ <!-- [if gte msEquation 12]>V(R)-V(4R)

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by
Alok Kumar Singh | Contributor-Level 10

a year ago

$σ_{1} = σ_{2}$

$\begin{array}{r} \frac{Q_{1}}{4 π R^{2}} = \frac{Q_{2} + Q_{1}}{4 π (16 R^{2})} \\ \Rightarrow Q_{1} + Q_{2} = 16 Q_{1} \\ \Rightarrow 1_{1} = Q_{2} \\ V (R) - V (4 R) \\ = (\frac{{K Q}_{1}}{R} + \frac{{K Q}_{2}}{4 R}) - (\frac{{K Q}_{1}}{4 R} + \frac{{K Q}_{2}}{4 R}) \\ = \frac{3 K Q Q_{1}}{4 R} = \frac{3 Q_{1}}{16 π \in R} \end{array}$

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New answer posted•a year ago

Electrostatics Class 12th

Two isolated conducting spheres S₁ and S₂ of radius (2/3)R and (1/3)R have 12µC and -3 µC charges, respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on S₁ and S₂ are respectively:

16 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by
Alok Kumar Singh | Contributor-Level 10

a year ago

When the two conducting spheres are connected by a wire, they will reach the same electric potential, V.
The total charge Q_total = 12µC + (-3µC) = 9µC. This total charge will redistribute.
Let the final charges be q? and q? + q? = 9µC.
The potential of a sphere is V = kq/r.
V? = V?
k q? /R? = k q? /R?

...Read more

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New answer posted•a year ago

Electrostatics Class 12th

In the circuit shown in the figure, the total charge is 750 µC and the voltage across capacitor C? is 20 V. Then the charge on capacitor C? is

12 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by
Alok Kumar Singh | Contributor-Level 10

a year ago

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New answer posted•a year ago

Electrostatics 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:

37 Views|Posted a year ago

Asked by Vishal Baghel

Answered by
Vishal Baghel | Contributor-Level 10

a year 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

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A two point charges 4 q and - q are fixed on the x -axis at x = - d 2 and x = d 2 , respectively. If a third point charge ' q ' is taken from the origin to x = d along the semicircle as shown in the figure, the energy of the charge will:

Concentric metallic hollow spheres of radii <!-- [if gte msEquation 12]>R R and 4 R <!-- [if gte msEquation 12]>4R Q 1 and Q 2 <!-- [if gte msEquation 12]>Q2 V ( R ) - V ( 4 R ) <!-- [if gte msEquation 12]>V(R)-V(4R)

Two isolated conducting spheres S₁ and S₂ of radius (2/3)R and (1/3)R have 12µC and -3 µC charges, respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on S₁ and S₂ are respectively:

In the circuit shown in the figure, the total charge is 750 µC and the voltage across capacitor C? is 20 V. Then the charge on capacitor C? is

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:

A two point charges $4 q$ and $- q$ are fixed on the $x$ -axis at $x = - \frac{d}{2}$ $\frac{}{}$ and $x = \frac{d}{2}$ $\frac{}{}$ , respectively. If a third point charge ' $q$ ' is taken from the origin to $x = d$ along the semicircle as shown in the figure, the energy of the charge will:

Concentric metallic hollow spheres of radii <!-- [if gte msEquation 12]>R $R$ and $4 R$ <!-- [if gte msEquation 12]>4R $Q_{1}$ and $Q_{2}$ <!-- [if gte msEquation 12]>Q2 $V (R) - V (4 R)$ <!-- [if gte msEquation 12]>V(R)-V(4R)