60 - Electricity Questions Answers

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

A solid conducting sphere of radius R is placed in a uniform electric field E as shown in figure-1.441. Due to electric field non uniform surface charges are induced on the surface of the sphere. Consider a point A on the surface of sphere at a polar angle θ from the direction of electric field as shown in figure. Find the surface density of induced charges at point A in terms of electric field and polar angle θ.

Asked By: DEBRAJ SAHA

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Joshi sir comment

To nullify the electric field inside, the conducting sphere creates a situation in which electric field inside due to charge on sphere would be same as that of given electric field. Now this is the case, which is same as Irodov 3.17.

See the video now

link:

Finally $E = \frac{σ_{0}}{3 ε_{0}} = \frac{σ}{3 ε_{0} \cos θ} N o w s o l v e$

Physics >> Electricity >> Current Electricity IIT JEE

Any method that can be used other than superposition principle?

Asked By: LUFFY

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Joshi sir comment

By superposition principle, it is solved in minimum time and effort, other method is very complex. So use superposition principle.

Divide current in point 1 and 2, considering symmetry throughout, we get the equivalent resistance of the whole network except resistor R between 1 and 2 is 2R

Now 2R and R in parallel between 1 and 2

Now solve

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

a small ball of mass 1kg and a charge 2/3 uC is placed at the center of a uniformly charged sphere of radius 1m and charge 1/3 mC. a narrow smooth horizontal groove is made in the sphere from centre to surface as shown in figure. the sphere is made to rotate about its vertical diameter at a constant rate of 1/2pi revolutions per second. find the spedd w.r.t ground with which the ball slide out from the groove. neglect any magnetic force acting on ball?

Asked By: VISHNU VARDHAN

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Joshi sir comment

$P o t e n t i a l d i f f e r e n c e f r o m c e n t r e t o s u r f a c e = 3 V / 2 - V = V / 2 h e r e V = k q / r (q i s c h a r g e o f s p h e r e) m {v_{r}}^{2} / 2 = V Q / 2 (Q i s c h a r g e o f p a r t i c l e) v_{r} i s r a d i a l v e l o c i t y B e s i d e s \tan g e n t i a l v e l o c i t y d u e t o r o t a t i o n o f s p h e r e = r ω = r 2 π n n o w s o l v e$

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

Asked By: BHASKAR JHA

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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 >> Current Electricity Medical Exam

if an area of isoscale right angle triangle(made of uniform wire of unit resistivity) is 8 unit. a battery of emf 10 unit is connected with across the longest branch then the current drawn from battery is approx ans is 3A

Asked By: AKANKSHA SINHA

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Joshi sir comment

area = l²/2 = 8 so l = 4

ρ = 1

so resistance of the 3 legs of the triangle = 4/A, 4/A, 4√2/A

battery is connected across the longest branch so eq. resistance = 4/A+4/A+4√2/A = 13.6/A

i think area of cross section will be given then calculate current

Physics >> Electricity >> Current Electricity IIT JEE

Can You explain some easy way to find potential between A and B?

Asked By: BONEY HAVELIWALA

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Joshi sir comment

use kirchhoff laws

4-10i+1-30x = 0

1-30x+20(i-x) = 0

solve it for i and x

potential across A and B = 4-10i+1 = 30x

60 - Electricity Questions Answers

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