Electrostatics

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New answer posted•a year ago

Electrostatics Class 12th

Two air filled capacitors are connected as shown. The left capacitor is suddenly filled with a dielectric of constant K. The work done by battery after that is $24 μ J .$ Dielectric constant K is:-

275 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$W D = Δ Q V$

$\Rightarrow Δ Q = 4 μ C$

$Q_{i} = \frac{2 \times 4}{2 + 4} \times 6 = 8 μ C$

$Q_{f} = 12 μ C = C_{e q} \times 6$

$C_{e q} = 2 μ F \Rightarrow C_{1} = C_{2} = 4 μ F$

$k = \frac{C_{1}}{C_{0}} = 2$

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New answer posted•a year ago

Electrostatics Class 12th

Three capacitors C₁ = $2 μ F, C_{2} = 12 μ F$ and C₃ = $12 μ F$ are connected as shown in figure. Find the ratio of the charges on capacitors C₁, C₂ and C₃ respectively:

35 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$Q_{1} = C_{1} V = 2 V μ C$

$Q_{2} = Q_{3} = \frac{C_{2} C_{3}}{C_{2} + C_{3}} V = \frac{6 \times 12}{6 + 12} V = 4 V μ C$

$Q_{1} : Q_{2} : Q_{3} = 1 : 2 : 2$

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New answer posted•a year ago

Electrostatics Class 12th

A capacitor is connected to a 20 V battery through a resistance of 10 $Ω .$ It is found that the potential difference across the capacitor rises to 2V in 1 $μ s .$ The capacitance of the capacitor is________ $μ F .$

Given $I n (\frac{10}{9}) = 0.105$

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

V = 10 $(1 ? e^{? t / R C})$ $^{}$

2 = 20 $(1 ? e^{? t / R C})$ $^{}$

$\frac{1}{10} = 1 ? e^{? t / R C}$

$e^{t / R C} = \frac{10}{9}$

$\frac{t}{R C} = l n (\frac{10}{9}) = 0.105$

$C = \frac{t}{R \times 0.105} = \frac{1 0^{? 6}}{10 \times 0.105} = 0.95 ? F$

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New answer posted•a year ago

Electrostatics Class 12th

A capacitor of capacitance 3µF is charged to a potential difference of 12V. Another identical capacitor, initially uncharged is filled with a dielectric having dielectric constant K and it is connected to the charged capacitor. The final common potential difference across both capacitors is found to be 2V. Find the value of k.

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

V_common = (C? V? + C? V? )/ (C? + C? )? 2 = (3? F)*12)/ (3? F)+ (3? F)*K)? 2 = 12/ (K+1)? K = 5

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New answer posted•a year ago

Electrostatics Class 12th

Two infinite line charges having linear charge density λ₁ and λ₂ are placed along the x and y axis respectively. The electric field at a point (x₁, y₁) in the first quadrant is inclined at an angle of 60° with the x-axis. Then which of the following relations is correct.

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

tan 60° = (2kλ? /y? )/ (2kλ? /x? )

(√3λ? /x? ) = (λ? /y? ); √3λ? y? = λ? x?

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New answer posted•a year ago

Electrostatics Class 12th

The diagram shows three concentric conducting spherical shells having radii $R, 2 R$ and $3 R$ . All three shells are given some charges. The initial potential of each shell is as mentioned in the figure. If the inner most shell is earthed then the final charge present on its outer surface would be equal to:

36 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

$V_{1} = \frac{k Q_{3}}{3 R} + \frac{k Q_{2}}{2 R} + \frac{k Q_{1}}{R} = 30$

$V_{2} = \frac{k Q_{3}}{3 R} + \frac{k Q_{2}}{2 R} + \frac{k Q_{1}}{2 R} = 15$

$V_{3} = \frac{k Q_{3}}{3 R} + \frac{k Q_{2}}{3 R} + \frac{k Q_{1}}{3 R} = 10$

Eq. (1) $&$ (2) $\to \frac{k Q_{1}}{R} = 30$

Eq. (2) $&$ (3) $\to \frac{k Q_{2}}{6 R} + \frac{k Q_{1}}{6 R} = 5 \Rightarrow \frac{k Q_{2}}{R} = - \frac{k Q_{1}}{R} + 30 = 0$

By (3) $\to \frac{k Q_{3}}{R} = 30 - \frac{k Q_{2}}{R} - \frac{k Q_{1}}{R} = 0$

Potential of innermost shell (earthed) $= \frac{k Q_{1}^{'}}{R} + \frac{k Q_{2}}{2 R} + \frac{k Q_{3}}{3 R} = 0$

$\Rightarrow \frac{k Q_{1}^{'}}{R} = 0 ∴ Q_{1}^{'} = 0$

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New answer posted•a year ago

Electrostatics Class 12th

Two capacitros are fully charged by a battery as shown in the figure.

If the capacitors are disconnected from battery and plate 1 is connected to plate 4 & plate 2 to plate 3, then find the potential difference (in volts) across capacitors finally.

15 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Charge initially on capacitor 1=10µC
Charge initially on capacitor 2=20µC
Finally the potential difference across both capacitor will be equal.
Let us assume that to be V.
1V + 2V = +10
V = 10 / 3 = 3.33V

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New answer posted•a year ago

Electrostatics Class 12th

A cube of side ‘a’ has point charges +Q located at each of its vertices except at the origin where the charge is -Q. The electric field at the centre of cube is:

14 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

If charge (-Q) at origin is replaced by (+Q), then electric field at the centre of the cube is zero. Thus, electric field at the centre of the cube is as if only (-2Q) charge is present at the origin.

$\vec{E} = \frac{1}{4 π ε_{0}} \frac{(- 2 Q)}{{(\frac{\sqrt[]{3}}{2} a)}^{2}} . \frac{1}{\sqrt[]{3}} (\hat{x} + \hat{y} + \hat{z}) = \frac{| - 2 Q}{3 \sqrt[]{3} π ε_{0} a^{2}} (\hat{x} + \hat{y} + \hat{z})$

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New answer posted•a year ago

Electrostatics Class 12th

Three capacitors each of capacitance C = 2μF are connected with a battery of emf 30V as shown in the figure. When the switch S is closed, the heat generated in circuit will be n/10 mJ. Find the value of n.

26 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Ui = ½C (V/3)² + ½C (V/3)² + ½C (2V/3)²
Uf = ½CV²
Wb = CV²/3
ΔH = [Wb - (Uf - Ui)] = CV²/6 = 0.30 mJ

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New answer posted•a year ago

Electrostatics Class 12th

Twenty seven drops of same size are charged at 220 V each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

11 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

(4/3)πR³ = 27 (4/3)πr³) ⇒ R = 3r (i)
v = Kq/r = V? /r = (q? /q? ) (r? /r? ) (i)
⇒ 220/V? = (q/ (27q) (3r/r) (i)
⇒ 220/V? = 1/9 (i)
⇒ V? = 220 × 9 = 1980 volt (i)

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