51 - Electric Field and Potential Questions Answers

Physics >> Electricity >> Electric Field and Potential IIT JEE

Asked By: BHASKAR JHA

is this question helpfull: 1 1 submit your answer

Joshi sir comment

$B y v o l u m e c o m p a r i s i o n r a d i u s o f c o m b i n e d d r o p = 2^{1 / 3} r n e t c h a r g e i n t h e n e w d r o p = 2 q n o w p u t t h e s e v a l u e s i n t h e f o r m u l a o f p o t e n t i a l$

Physics >> Electricity >> Electric Field and Potential IIT JEE

Initially the spheres a and b are at potential be a and b respectively now sphere b is earthd by closing the switch the poential of a will become

Asked By: ROMEO ROMI

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Joshi sir comment

No meaning can be incurred by this language. Rectify it for getting answer

Physics >> Electricity >> Electric Field and Potential IIT JEE

Asked By: DARIUS

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Joshi sir comment

$S i n c e t h i c k n e s s i s s a m e s o f o r c e p e r u n i t l e n g t h s h o u l d b e s a m e F o r c e p e r u n i t l e n g t h = F / 2 π R C h a r g e d e n s i t y i n t h e s p h e r e i s p r o p o r t i o n a l t o E s o F α q E α σ A E α E^{2} R^{2} T h u s f o r c e p e r u n i t l e n g t h α E^{2} R^{2} / 2 π R α E^{2} R N o w c o m p a r e t h e t w o c a s e s$

Physics >> Electricity >> Electric Field and Potential IIT JEE

Asked By: KRISHAV RAJ SINGH

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Solution by Joshi sir

$\frac{k q_{1} q_{2}}{{l_{3}}^{2}} = \frac{k q_{2} q_{3}}{{l_{1}}^{2}} = \frac{k q_{3} q_{1}}{{l_{2}}^{2}} l_{1} + l_{2} + l_{3} = L f r o m t h e s e 2 e q u a t i o n s \frac{L \frac{1}{\sqrt{q_{1}}}}{\frac{1}{\sqrt{q_{1}}} + \frac{1}{\sqrt{q_{2}}} + \frac{1}{\sqrt{q_{3}}}}, \frac{L \frac{1}{\sqrt{q_{2}}}}{\frac{1}{\sqrt{q_{1}}} + \frac{1}{\sqrt{q_{2}}} + \frac{1}{\sqrt{q_{3}}}}, \frac{L \frac{1}{\sqrt{q_{3}}}}{\frac{1}{\sqrt{q_{1}}} + \frac{1}{\sqrt{q_{2}}} + \frac{1}{\sqrt{q_{3}}}}$

Physics >> Electricity >> Electric Field and Potential IIT JEE

A particle a having a charge of $5.0 \times 10^{- 7} C$ is fixed in. a vertical wall. A second particle B of mass 100 g and. having equal charge is supended by a silk thread. of length 30 cm form the wall. The point of suspension is. 30 cm above the particle A. Find the angle of the thread. with the vertical when it stays in equilibrium.

Asked By: UDDESH KUMAR SABAT

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Solution by Joshi sir

Physics >> Electricity >> Electric Field and Potential AIEEE

A Volt-meter of 20000 Ohm resistance is connected with R ohm resistance in series. if we apply 110 V of pd to the mechanism, the Volt meter shows 5 V reading.. so find the R...

Asked By: BONEY HAVELIWALA

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Joshi sir comment

current on both the resistances will be same

so V₁/R₁ = V₂/R₂

5/20000 = 105/R

solve

Physics >> Electricity >> Electric Field and Potential Medical Exam

An isolated conducting sphere of radius r has given charge q then PE??

ans is q²/8πe₀r

But Sir ,why the ans q²/2πe₀r is incorrect??

Asked By: SARIKA SHARMA

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Joshi sir comment

let we have given x charge and want to give dx also to the sphere

so dE = [1/4πε₀]xdx/r

now integrate and take limits of x as 0 to q

Physics >> Electricity >> Electric Field and Potential Medical Exam

2 Equal -ve charges -q r placed at pt(0,a) and (o,-a) on y axis , one +ve charge +q at rest is left from pt(2a,0). this charge will not execute SHM . WHY?? even when F α -x condition is satisfying as force is due to both charges is acting & motion is also opposing during oscillating

Asked By: SARIKA SHARMA

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Joshi sir comment

Solve the question for force, you will get a condition according to which yours statement "even when F α -x condition is satisfying"

is incorrect

Physics >> Electricity >> Electric Field and Potential Medical Exam

Two identical particles each of mass M & charge Q are placed some distance apart . If they are in equilm under mutual gravitational & electric force then calculate the order of signature of charge Q/M in SI system.

ANS = 10^-10 C/kg

Asked By: SARIKA SHARMA

is this question helpfull: 4 0 read solutions ( 1 ) | submit your answer

Joshi sir comment

on comparing gravitational force and electrical force,

we get 9*10⁹ * q²/r² = 6.67*10^-11* m²/r²

so q/m = 10^-10approx

Physics >> Electricity >> Electric Field and Potential IIT JEE

a fixed uniformly charged ring of radius 3m has a positive linear charge density 50/3µC/m. a point charge 5μC is moving towards the ring along its axis such that its kinetic energy 4m away is 5J. Its kinetic energy at the centre of the ring will be?

Asked By: ROCKANDROLL

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Joshi sir comment

compare total energy at the two points and get answer

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