Tags Physics

Physics

Get insights from 5.6k questions on Physics, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics

Follow Ask Question

5.6k

Questions

Discussions

Active Users

Followers

New question posted

a year ago

Physics Class 12th

sjfkjffWNEFLWENFOILW

0 Follower 1 View

New question posted

a year ago

Physics Science

How can I get the books for Physics, Chemistry and Mathematics section of MHT CET 2026?

0 Follower 9 Views

New answer posted

a year ago

Physics Electrostatic Potential and Capacitance

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

$(E_{2}$ - $E_{1}) . \hat{n} = \frac{σ}{ε_{0}}$

Where $\hat{n}$ is a unit vector normal to the surface at a point and s is the surface charge density at that point. (The direction of $\hat{n}$ is from side 1 to side 2.) Hence, show that just outside a conductor, the electric field is $σ \hat{n}$ / $ε_{0}$ .

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

[Hint: For (a), use Gauss's law. For, (b) use the fact that work done by electrostatic field on a closed lo

...more

(a) Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by

$(E_{2}$ - $E_{1}) . \hat{n} = \frac{σ}{ε_{0}}$

(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.

[Hint: For (a), use Gauss's law. For, (b) use the fact that work done by electrostatic field on a closed loop is zero.]

less

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

2.16 Electric field on one side of a charged body is $E_{1}$ and electric field on the other side of the same body is $E_{2} .$ If infinite plane charged body has a uniform thickness, then electric field due to one surface of the charged body is given by,

$\overset{?}{E_{1}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(i)

Where,

$\overset{?}{n}$ = unit vector normal to the surface at a point

$σ$ = surface charge density at that point

Electric field due to the other surface of the charged body is given by

$\overset{?}{E_{2}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(ii)

$E l e c t r i c f i e l d a t a n y p o i n t$ due to the two surfaces,

$\overset{?}{E_{2}} - \overset{?}{E_{1}} = \frac{σ}{2 ε_{0}} \overset{?}{n} + \frac{σ}{2 ε_{0}} \overset{?}{n} = \frac{σ}{ε_{0}} \overset{?}{n}$ ……(iii)

$S i n c e i n s i d e a c l o s e d c o n d u c t o r, \overset{?}{E_{1}} = 0,$

$\overset{?}{E}$ = $\overset{?}{E_{2}}$ = $\frac{σ}{ε_{0}} \overset{?}{n}$

Therefore, the electric field just outside the conductor is $\frac{σ}{ε_{0}} \overset{?}{n}$

...more

$\overset{?}{E_{1}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(i)

Where,

$\overset{?}{n}$ = unit vector normal to the surface at a point

$σ$ = surface charge density at that point

Electric field due to the other surface of the charged body is given by

$\overset{?}{E_{2}}$ = $\frac{σ}{2 ε_{0}} \overset{?}{n}$ ………….(ii)

$E l e c t r i c f i e l d a t a n y p o i n t$ due to the two surfaces,

$\overset{?}{E_{2}} - \overset{?}{E_{1}} = \frac{σ}{2 ε_{0}} \overset{?}{n} + \frac{σ}{2 ε_{0}} \overset{?}{n} = \frac{σ}{ε_{0}} \overset{?}{n}$ ……(iii)

$S i n c e i n s i d e a c l o s e d c o n d u c t o r, \overset{?}{E_{1}} = 0,$

$\overset{?}{E}$ = $\overset{?}{E_{2}}$ = $\frac{σ}{ε_{0}} \overset{?}{n}$

Therefore, the electric field just outside the conductor is $\frac{σ}{ε_{0}} \overset{?}{n}$

When a charged particle is moved from one point to the other on a closed loop, the work done by the electrostatic field is zero. Hence, the tangential component of electrostatic field is continuous from one side of a charged surface to the other.

less

0 0

New answer posted

a year ago

Physics Electrostatic Potential and Capacitance

2.31 (a) Two large conducting spheres carrying charges $Q_{1}$ and $Q_{2}$ are brought close to each other. Is the magnitude of electrostatic force between them exactly given by $Q_{1} Q_{2}$ /4 $π ε_{0} r^{2}$ , where r is the distance between their centres?

(b) If Coulomb's law involved 1/ $r^{3}$ dependence (instead of 1/ $r^{2}$ ), would Gauss's law be still true ?

(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if t

...more

(b) If Coulomb's law involved 1/ $r^{3}$ dependence (instead of 1/ $r^{2}$ ), would Gauss's law be still true ?

(c) A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?

(d) What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?

(e) We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?

(f) What meaning would you give to the capacitance of a single conductor?

(g) Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

less

0 Follower 8 Views

alok kumar singh

Contributor-Level 10

2.31 The force between two conducting spheres is not exactly given by the expression $Q_{1} Q_{2}$ /4 $? ?_{0} r^{2}$ , because there is non-uniform charge distribution on the spheres.

Gauss's law will not be true, if Coulomb's law involved $\frac{1}{r^{3}}$ dependence, instead of $\frac{1}{r^{2}}$ , on r

Yes. If a small test charge is released at rest at a point in an electrostatic field configuration, then it will travel along the field lines passing through the point, only if the field lines are straight. This is because the field lines give the direction of acceleration and not the velocity.

Whenever the electron completes an orbit, either circular or elli

...more

2.31 The force between two conducting spheres is not exactly given by the expression $Q_{1} Q_{2}$ /4 $? ?_{0} r^{2}$ , because there is non-uniform charge distribution on the spheres.

Gauss's law will not be true, if Coulomb's law involved $\frac{1}{r^{3}}$ dependence, instead of $\frac{1}{r^{2}}$ , on r

Whenever the electron completes an orbit, either circular or elliptical, the work done by the field of a nucleus is zero.

No. Electric field is discontinuous across the surface of a charged conductor. However, electric potential is continuous.

The capacitance of a single conductor is considered as a parallel plate capacitor with one of its two plates at infinity.

Water has an unsymmetrical space as compared to mica. Since it has a permanent dipole moment, it has a greater dielectric constant than mica.

less

0 0

New question posted

a year ago

Physics Science

What is the maximum time one can use for Physics and Mathematics in MHT-CET?

0 Follower 4 Views

New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Five

A helicopter of mass 2000 kg rises with a vertical acceleration of 15 m s^–2. The total mass of the crew and passengers is 500 kg. Give the magnitude and direction of the (g = 10 m s^–2)

(a) Force on the floor of the helicopter by the crew and passengers.

(b) Action of the rotor of the helicopter on the surrounding air.

(c) Force on the helicopter due to the surrounding air.

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation- mass of helicopter m =2000kg

Mass of the crew and passengers m= 500kg

Acceleration a =15m/s² and g = 10m/s²

a) force on the floor of the helicopter by the crew and passengers

m (g+a)= 500 (10+15)N= 12500N

b) action of rotor of the helicopter on the surroundings air= (m₁+m₂) (g+a)

= (2000+500) (10+15)= 2500 (25)= 62500N

c) force on the helicopter due to surroundings air

= reaction of force applied by helicopter

= 62500N

0 0

New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Five

A rectangular box lies on a rough inclined surface. The co-efficient of friction between the surface and the box is µ. Let the mass of the box be m.

(a) At what angle of inclination θ of the plane to the horizontal will the box just start to slide down the plane?

(b) What is the force acting on the box down the plane, if the angle of inclination of the plane is increased to α θ > ?

(c) What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed?

(d) What is the force needed to be applied upwards along the plane to make the box move up the plane with

...more

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation -for the box to just starts sliding down mg

sin $θ = t = μ N = μ m g c o s θ$

0 0

New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Five

There are four forces acting at a point P produced by strings as shown in Fig. , which is at rest. Find the forces F₁ and F₂

0 Follower 8 Views

alok kumar singh

Contributor-Level 10

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation- on resolving forces into rectangular components in equilibrium forces (F₁+1/ $\sqrt[]{2}$ )N are equal to $\sqrt[]{2} N$ and F₂ is equal to ( $\sqrt[]{2} + 1 / \sqrt[]{2}$ )N

F₁+1/ $\sqrt[]{2}$ = $\sqrt[]{2}$

F₁= $\sqrt[]{2} - 1 / \sqrt[]{2}$ = $\frac{2 - 1}{\sqrt[]{2}} = \frac{1}{\sqrt[]{2}} = 0.707 N$

F₂= $\sqrt[]{2} + \frac{1}{\sqrt[]{\sqrt[]{2}}} = \frac{2 + 1}{\sqrt[]{2}} = \frac{3}{\sqrt[]{2}} = 2.121 N$

0 0

New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Five

A cricket bowler releases the ball in two different ways

(a) Giving it only horizontal velocity, and

(b) Giving it horizontal velocity and a small downward velocity.

The speed vs at the time of release is the same. Both are released at a height H from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.

0 Follower 9 Views

alok kumar singh

Contributor-Level 10

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation – horizontal velocity u_x= v_s

During projectile motion horizontal velocity remains unchanged

V_x=u_x=v_s

In vertical direction v_y²= u_y²+2gH

0 0

New answer posted

a year ago

Physics Physics NCERT Exemplar Solutions Class 11th Chapter Five

The displacement vector of a particle of mass m is given by r(t)=ȋ A cos $w t +$ ĵ B sinwt.

(a) Show that the trajectory is an ellipse.

(b) Show that F= -mw²/r

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation- displacement vector of particle is r (t)=? Acos $w t +$ ? Bsinwt

X=Acoswt

x/A= coswt………1

displacement along y axis is

y=Bsinwt

y/B= sinwt……….2

squaring and then adding eqn1 and 2 we get

x²/A²+y²/B²=cos²wt+sin²wt =1

this is an equation of ellipse. Therefore trajectory of particle is an ellipse.

b)v= dr/dt= id/dt (Acoswt)+jd/dt (Bsinwt)

? [A (-sinwt)w]+? [B (coswt).w]

= -? Awsinwt+? Bwcoswt

Acceleration a= dv/dt

So a= -? Awd/dt (sinwt)+? Bw [-sinwt]w

=? Aw²coswt-? Bsinwt

= -w²r

So force acting on the particle f=ma=-mrw²

0 0

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

On Shiksha, get access to

66k Colleges
1.2k Exams
705k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Physics

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience