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Class 12th physics ncert exemplar solution class 12th chapter one

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

Explanation- as we know F= Qq/r²= 1dyne = 1esu of charge²/1cm²

1 esu of charge = M^1/2L^3/2T^—1

Q=xC where x is dimensionless quantity

So F= $Q + q_{2} = q_{1}$ =1 dyne = 10^-5N

Taking x= 1/3 $Q = q_{1} - q_{2}$ ⁹

After solving we get 1/4 $q_{1} + q_{2}$

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Two fixed, identical conducting plates (α &β ) , each of surface area S are charged to –Q and q, respectively, where Q > q > 0. A third identical plate (γ ), free to move is located on the other side of the plate with charge q at a distance d . The third plate is released and collides with the plate β . Assume the collision is elastic and the time of collision is sufficient to redistribute charge amongst β &γ . (a) Find the electric field acting on the plate γ before collision. (b) Find the charges on β and γ after the collision.

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Class 12th physics ncert exemplar solution class 12th chapter one

Contributor-Level 10

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

Explanation- net electric field at plate γ due to two other plate

From plate 1 ,E₁= $\frac{- Q}{S (2 ε_{0})} t o t h e l e f t$

From plate 2 ,E₂= $\frac{q}{S (2 ε_{0})} t o t h e r i g h t$

Total electric field E= E₁+ E₂ = $\frac{q - Q}{S (2 ε_{0})}$ to the left , if Q>q

electric field at o due to plate α = $\frac{- Q}{S (2 ε_{0})} t o t h e l e f t$

electric field at o due to plate β = $\frac{q 1}{S (2 ε_{0})} t o t h e r i g h t$

electric field at o due to plate γ = $\frac{q 2}{S (2 ε_{0})} t o t h e l e f t$

as the electric field at o is zero therefore

As there is no loss of charge on collision

Q+q= $\frac{Q + q_{2}}{S (2 ε_{0})} = \frac{q_{1}}{S (2 ε_{0})}$

On solving these

q₁= (Q+q/2)= charge on plate β

q₂= (q/2)= charge on plate γ

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Consider a sphere of radius R with charge density distributed as ρ (r ) = kr for r ≤ R for r > R .

(a) Find the electric field at all points r.

(b) Suppose the total charge on the sphere is 2e where e is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.

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Class 12th physics ncert exemplar solution class 12th chapter one

Contributor-Level 10

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

Explanation-

(i) when r

$? E . d s$ = $\frac{1}{ε_{0}} \int ρ d V$ and we know that V = $\frac{4}{3} π r^{3}$

$d V = 3 * \frac{4}{3} π r^{3} d r$ = 4 $π r^{2} d r$

$? E . d s$ = $\frac{1}{ε_{0}} 4 π K \int r^{3} d r$

E(4 ) $π r^{2}$ = $\frac{4 π K r^{4}}{ε_{0} 4}$ = $\frac{k r^{2}}{4 ε_{0}}$ so it is clear that E is radially outwards.

When r>R

$? E . d s$ = $\frac{1}{ε_{0}} \int ρ d V$

E= $\frac{k R^{4}}{4 ε_{0} r^{2}}$ again field is outwards

(ii) When two protons are there then they must be on opposite sides or we can say along the end of diameter

So q= $\int ρ d V = \int K r 4 π r^{2} d r$

.q= 4 $π K \frac{R^{4}}{4} = 2 e, s o K = \frac{2 e}{π R^{4}}$

If protons 1 and 2 are embedded at distance r from the center of the sphere then force will be

F=eE=- $\frac{e K r^{2}}{4 ε_{0}}$ but the force applied by proton F= $\frac{e^{2}}{4 π ε_{0} ({2 r)}^{2}}$

By adding these F= - $\frac{e K r^{2}}{4 ε_{0}} + \frac{e^{2}}{4 π ε_{0} ({2 r)}^{2}}$

If F=0 then $\frac{e K r^{2}}{4 ε_{0}} = \frac{e^{2}}{4 π ε_{0} ({2 r)}^{2}}$ &nbs

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

Explanation-

(i) when r

$? E . d s$ = $\frac{1}{ε_{0}} \int ρ d V$ and we know that V = $\frac{4}{3} π r^{3}$

$d V = 3 * \frac{4}{3} π r^{3} d r$ = 4 $π r^{2} d r$

$? E . d s$ = $\frac{1}{ε_{0}} 4 π K \int r^{3} d r$

E(4 ) $π r^{2}$ = $\frac{4 π K r^{4}}{ε_{0} 4}$ = $\frac{k r^{2}}{4 ε_{0}}$ so it is clear that E is radially outwards.

When r>R

$? E . d s$ = $\frac{1}{ε_{0}} \int ρ d V$

E= $\frac{k R^{4}}{4 ε_{0} r^{2}}$ again field is outwards

(ii) When two protons are there then they must be on opposite sides or we can say along the end of diameter

So q= $\int ρ d V = \int K r 4 π r^{2} d r$

.q= 4 $π K \frac{R^{4}}{4} = 2 e, s o K = \frac{2 e}{π R^{4}}$

If protons 1 and 2 are embedded at distance r from the center of the sphere then force will be

F=eE=- $\frac{e K r^{2}}{4 ε_{0}}$ but the force applied by proton F= $\frac{e^{2}}{4 π ε_{0} ({2 r)}^{2}}$

By adding these F= - $\frac{e K r^{2}}{4 ε_{0}} + \frac{e^{2}}{4 π ε_{0} ({2 r)}^{2}}$

If F=0 then $\frac{e K r^{2}}{4 ε_{0}} = \frac{e^{2}}{4 π ε_{0} ({2 r)}^{2}}$ so after solving r= $\frac{R}{8^{1 / 4}}$

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In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained if matter carried a net charge. Suppose that the Universe is made up of hydrogen atoms with a number density N, which is maintained a constant. Let the charge on the proton be: ep=−(1+y)e where e is the electronic charge. Find the critical value of y such that expansion may start

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Class 12th physics ncert exemplar solution class 12th chapter one

Contributor-Level 10

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

Explanation- let us consider that universe is of radius R

And we know that hydrogen is made up one electron and proton so net charge is

-(1+y)e+e = -ye

Now the number of hydrogen atom in the universe = N $* \frac{4}{3} π R^{3}$

So total charge is = -ye $*$ N $* \frac{4}{3} π R^{3}$

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

Explanation- let us consider that universe is of radius R

And we know that hydrogen is made up one electron and proton so net charge is

-(1+y)e+e = -ye

Now the number of hydrogen atom in the universe = N $* \frac{4}{3} π R^{3}$

So total charge is = -ye $*$ N $* \frac{4}{3} π R^{3}$

According to gauss theorem $? E . d s = \frac{q}{ε_{0}}$

So E $* 4 π R^{2} = \frac{- y e * N * \frac{4}{3} π R^{3}}{ε_{0}}$

E= -NRye/3 $ε_{0}$

And we know electrostatic force F=qE = -ye(-NRye/3 ) $ε_{0}$ by using the value of E and q

F= $\frac{y^{2} e^{2} N R}{3 ε_{0}}$ force is positive which shows that force is repulsive.

But the gravity is g= $\frac{G M}{R^{2}}$ and gravitational force is F= $\frac{G M m}{R^{2}}$

But M= N $* \frac{4}{3} π R^{3} m_{p}$ where $m_{p}$ is mass of proton

So using this mass in above eqn so force = $\frac{G N * \frac{4}{3} π R^{3} m_{p} m}{R^{2}}$

But her masses of proton and hydrogen are equal so take as m

F= $\frac{G N * 4 π R m^{2}}{3}$

So calculate its critical value electrostatic force and gravitational force both are equal.

So, $\frac{G N * 4 π R m^{2}}{3} = \frac{y^{2} e^{2} N R}{3 ε_{0}}$

So y²= 4 $π ε_{0} \frac{m^{2}}{e^{2}}$ after using standard values we get value of y= approx. 10^-18

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Five charges,q each are placed at the comers of a regular pentagon of side a.

i) What will be the electric field at O, the centre of the pentagon?
(ii) What will be the electric field at O if the charge from one of the comers (say A) is removed?
(iii) What will be the electric field at O if the charge q at A is replaced by – q?
(b) How would your answer to (a) be affected if pentagon is replaced by n-sided regular polygon with charge q at each of its comers?

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Class 12th physics ncert exemplar solution class 12th chapter one

Contributor-Level 10

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

Explanation- (a)

(i) The electric field at the center of pentagon is zero because the distance from the center is same.

(ii) The field through one charge is Kq/r²

(iii) When one charge is positive and other is negative then net force towards negative charge. So net force is Kq/r²+ Kq/r²= 2Kq/r²

(b) It doesn't depend upon the number of sides increasing the net electric field is zero.

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Five charges,q each are placed at the comers of a regular pentagon of side a.

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Can I download the NCERT Exemplar PDF for Electrostatic Potential and Capacitance for offline use?

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Contributor-Level 10

Yes, students can download the NCERT Exemplar PDF for Class 12 Physics Chapter 2. This offers various types of conceptual and application-based questions. Students can practice these offline at their own schedule. Practicing from PDF helps in improving problem-solving skills.

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Are NCERT Exemplar Class 12 Physics Chapter 2 solutions sufficient for exam preparation?

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Preparing from the NCERT Exemplar Chapter 2 solutions will help in providing a strong concept foundation as it covers a wide range of question types from basic to advanced. If one practices the exemplar, they can score high in Board exams and entrance exams. However, it is advisable to first start from the NCERT textbook for the theory of the concepts and later on practice exemplars to boost the understanding and solve various types of questions based on these concepts. The exemplars are aligned with the CBSE syllabus and offer great study material for strengthening concept understanding and improving problem-solving accuracy.

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What is the significance of solving NCERT Exemplar problems for Class 12 Physics Chapter 2?

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NCERT Exemplar problems are designed by the subject matter experts. It helps in deepening conceptual understanding and improving the application skills of the students. Preparing the NCERT exemplar Chapter 2 helps students in mastering complex topics like capacitors, electric potential, and energy storage which are important for the board exam preparations and entrance tests like JEE and NEET.

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Are NCERT Exemplar Class 12 Physics Chapter 2 solutions sufficient for exam preparation?

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