Tags Electrostatic Potential and Capacitance

Electrostatic Potential and Capacitance

Get insights from 125 questions on Electrostatic Potential and Capacitance, answered by students, alumni, and experts. You may also ask and answer any question you like about Electrostatic Potential and Capacitance

Follow Ask Question

125

Questions

Discussions

Active Users

Followers

New answer posted

2 months ago

Electrostatic Potential and Capacitance Class 12th

Two identical positive charge Q each are fixed at a distance of '2a' apart from each other. Another point charge q₀ with mass 'm' is placed at midpoint between two fixed charges. For a small displacement along the line joining the fixed charges, the charge q₀ executes SHM. The time period of oscillation of charge q₀ will be:

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

$F_{R} = {\frac{k Q q_{0}}{{(a - x)}^{2}} - \frac{k Q q_{0}}{{(a + x)}^{2}}}$

$= - k Q q_{0} {\frac{a^{2} + x^{2} + 2 a x - a^{2} - x^{2} + 2 a x}{(a^{2} - x^{2})}}$

$F_{R} = - \frac{k Q q_{0} (4 a x)}{{(a^{2} - x^{2})}^{2}}$

$T = 2 π \sqrt[]{\frac{a^{3} m}{4 k Q q_{0}} = \sqrt[]{\frac{4 π^{2} a^{3} m * 4 π ε_{0}}{4 Q q_{0}}}} = \sqrt[]{\frac{4 π^{3} ε_{0} m a^{3}}{Q q_{0}}}$ .

0 0

New answer posted

2 months ago

Electrostatic Potential and Capacitance Class 12th

Two identical thin metal plates has charge q₁ and q₂ respectively such that q₁ > q₂. The plates were brought close to each other to from a parallel plate capacitor of capacitance C. The potential difference between them is

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

V = $\frac{q_{(c a p a c i t a n c e)}}{C} = \frac{q_{1} - q_{2}}{2 c}$

0 0

New answer posted

2 months ago

Electrostatic Potential and Capacitance Class 12th

Two identical metallic spheres A and B when placed at certain distance in air repel each other with force of F. Another identical uncharged sphere C is first place in contact with A and then in contact with B and finally placed at midpoint between sphere A and B. The force experienced by sphere C will be

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

For sphere 'C' after contacting with 'A'. q_A = q_C = $\frac{q}{2} .$

offer contacting with 'B'. q_B = q_C = $\frac{3 q}{4}$

$F_{N e t} = | F_{1} - F_{2} |$

F₂ = $\frac{K \cdot 3 q^{2} * 4}{r^{2} 8} = \frac{3}{2} \frac{k q^{2}}{r^{2}}$

$F_{1} = \frac{9 k q^{2} * 4}{16 r^{2}} = \frac{9 k q^{2}}{4 r^{2}} = \frac{9}{4} F$

$∴ F_{N e t} = | \frac{9}{4} F - \frac{3}{2} F |$

= $\frac{9 - 6}{4} F = \frac{3}{4} F$

0 0

New answer posted

2 months ago

Electrostatic Potential and Capacitance Class 12th

A capacitor of capacitance 50 pF is charged by 100 V source. It is then connected to another uncharged identical capacitor. Electrostatic energy loss in the process is _________nJ.

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

Q = CV

= 50 * 10^-12 * 100

= 5 * 10^-12 * 10³

Q = 5 * 10^-9C

U_i (Initial Energy)

= $\frac{Q^{2}}{2 C} = \frac{25 * 1 0^{- 18}}{2 * 50 * 1 0^{- 12}}$

= $\frac{1}{4} * 1 0^{- 6}$

Now capacitor is connected to an identical uncharged capacitor

kvL

$\frac{Q_{1}}{C} = \frac{Q_{2}}{C} = 0$

q₁ = q₂

Initial change Q₁ + Q₂ = Q

$\Rightarrow 2 Q_{1} = Q$

$Q_{1} = \frac{Q}{2}$

$u_{F} = \frac{Q_{1}^{2}}{2 C} + \frac{Q_{1}^{2}}{2 C}$

$= \frac{{(\frac{Q}{2})}^{2}}{2 C} + \frac{{(Q / 2)}^{2}}{2 C}$

$= 2 * \frac{Q^{2}}{4 * 2 C}$

...more

Q = CV

= 50 * 10^-12 * 100

= 5 * 10^-12 * 10³

Q = 5 * 10^-9C

U_i (Initial Energy)

= $\frac{Q^{2}}{2 C} = \frac{25 * 1 0^{- 18}}{2 * 50 * 1 0^{- 12}}$

= $\frac{1}{4} * 1 0^{- 6}$

Now capacitor is connected to an identical uncharged capacitor

kvL

$\frac{Q_{1}}{C} = \frac{Q_{2}}{C} = 0$

q₁ = q₂

Initial change Q₁ + Q₂ = Q

$\Rightarrow 2 Q_{1} = Q$

$Q_{1} = \frac{Q}{2}$

$u_{F} = \frac{Q_{1}^{2}}{2 C} + \frac{Q_{1}^{2}}{2 C}$

$= \frac{{(\frac{Q}{2})}^{2}}{2 C} + \frac{{(Q / 2)}^{2}}{2 C}$

$= 2 * \frac{Q^{2}}{4 * 2 C}$

$U_{f} = \frac{Q_{2}}{4 * 2 C}$

Loss in energy = U_i = U_t

= $\frac{Q^{2}}{2 C} - \frac{Q^{2}}{4 C}$

= $\frac{Q^{2}}{4 C}$

= $\frac{1}{2} [\frac{Q^{2}}{2 C}]$

$\frac{1}{2} * 0.25 * 1 0^{- 6}$

= 125 * 10^-9

= 125 nJ

less

0 0

New answer posted

2 months ago

Electrostatic Potential and Capacitance Class 12th

A force of 10 N acts on a charged particle placed between two plates of a charged capacitor. If one plate of capacitor is removed, then the force acting on that particle will be.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

$F = q E [E = \frac{σ}{ε_{0}}]$

$F = \frac{q σ}{\in_{0}} = 10 N$

Final case – Now one plate is removed, Electric field will becomes half; So, force will become half.

$F' = \frac{10}{2} = 5 N$

0 0

New answer posted

2 months ago

Electrostatic Potential and Capacitance Class 12th

In the figure, potential difference between $A$ and $B$ is:

0 Follower 1 View

alok kumar singh

Contributor-Level 10

In forward bias diode act as a short circuit wire. Hence, the equivalent circuit is now.

So, $_{} \frac{}{}$ $V_{a b} = \frac{30}{5 + 10} * 5 = 10 V$

$50.00$

0 0

New answer posted

2 months ago

Electrostatic Potential and Capacitance Class 12th

A condenser of $2 μ F$ capacitance is charged steadily from 0 to 5 C. Which of the following graph represents correctly the variation of potential difference (V) across it's plates with respect to the charge (Q) on the condenser?

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

Q = CV

$V = \frac{Q}{C}$

Straight line with slope = $\frac{1}{C ?}$ $\frac{}{}$

$S l o p e = \frac{1}{C} = \frac{1}{2 * 1 0^{- 6}} = 5 * 1 0^{5}$

0 0

New answer posted

2 months ago

Electrostatic Potential and Capacitance Class 12th

Which of the following physical quantities have the same dimensions?

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

Electric Displacement

$\begin{array}{l} \vec{D} = \in o \vec{E} \\ [D] = [\in_{0} \vec{E}] = [\in_{0} \frac{σ}{\in_{0}}] \\ [D] = [σ] \end{array}$

Electric displacement $(\vec{D})$ and surface change density

0 0

New answer posted

2 months ago

Electrostatic Potential and Capacitance Class 12th

Two capacitors, each having capacitance $40 μ F$ are connected in series. The space between one of the capacitors is filled with dielectric material of constant K such that the equivalence capacitance of the system became $24 μ F$ . The value of K will be:

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

$C_{e q} = \frac{C_{1} C_{2}}{C_{1} + C_{2}} = \frac{40 * 40 k}{40 (1 + k)} = 24$

$\begin{array}{l} \Rightarrow 40 k = 24 + 24 k \\ \Rightarrow 16 k = 24 \end{array}$

$k = \frac{24}{16} = \frac{3}{2} = 1.5$

0 0

New answer posted

2 months ago

Electrostatic Potential and Capacitance Class 12th

In the given figure, the value of V₀ will be ________ V.

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

K V L Loop 1

6 - l * 1 -l₁ * 1 - 4 = 0

2 = l + l₁ - (i)

K V L Loop 2

4 + l₁ * 1 - (l – l₁) * 1 – 2 = 0

$\Rightarrow 2 + l_{1} - l = 0$

l = 2 + 2l₁ &nb

...more

K V L Loop 1

6 - l * 1 -l₁ * 1 - 4 = 0

2 = l + l₁ - (i)

K V L Loop 2

4 + l₁ * 1 - (l – l₁) * 1 – 2 = 0

$\Rightarrow 2 + l_{1} - l = 0$

l = 2 + 2l₁ - (ii)

from (i) & (ii)

l₁ = 0 v₀ = 4

less

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Electrostatic Potential and Capacitance

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience