NCERT Exemplar Class 12 Physics Chapter 1 Electric Charges and Fields

1. 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?

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.

          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 $\times \frac{4}{3}\pi R^3$

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

             According to gauss theorem $\oint E.ds=\frac{q}{{\epsilon }_0}$

              So E $\times 4\pi R^2=\frac{-\mathrm{y}\mathrm{e}\times \mathrm{}\mathrm{N}\times \frac{4}{3}\pi R^3}{{\epsilon }_0}$

              E= -NRye/3 ${\epsilon }_0$

             And we know electrostatic force F=qE = -ye(-NRye/3 ) ${\epsilon }_0$ by using the value of E and q

             F= $\frac{y^2{e}^{2}NR}{3{\epsilon }_0}$ force is positive which shows that force is repulsive.

             But the gravity is g= $\frac{GM}{R^2}$ and gravitational force is F= $\frac{GMm}{R^2}$

             But M= N $\times \frac{4}{3}\pi R^3{m}_p$  where $m_p$ is mass of proton

             So using this mass in above eqn so force = $\frac{G\mathrm{N}\times \frac{4}{3}\pi R^3{m}_{p}m}{R^2}$

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

             F= $\frac{G\mathrm{N}\times 4\pi R{m}^2}{3}$

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

             So, $\frac{G\mathrm{N}\times 4\pi R{m}^2}{3}=\frac{y^2{e}^{2}NR}{3{\epsilon }_0}$  

             So y²= 4 $\pi {\epsilon }_0\frac{m^2}{e^2}$  after using standard values we get value of y= approx. 10^-18

3. 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.

Explanation-

(i) when r

$\oint E.ds$   = $\frac{1}{{\epsilon }_0}\int \mathrm{\rho }\mathrm{d}\mathrm{V}$ and we know that V = $\frac{4}{3}\pi r^3$

$dV=3\times \frac{4}{3}\pi r^{3}dr$ = 4 $\pi r^{2}dr$

  $\oint E.ds$ = $\frac{1}{{\epsilon }_0}4\pi K\int r^{3}dr$

  E(4 ) $\pi r^2$ = $\frac{4\pi K{r}^4}{{\epsilon }_{0}4}$ = $\frac{k{r}^2}{4{\epsilon }_0}$ so it is clear that E is radially outwards.

  When r>R

$\oint E.ds$ = $\frac{1}{{\epsilon }_0}\int \mathrm{\rho }\mathrm{d}\mathrm{V}$

              E= $\frac{k{R}^4}{4{\epsilon }_0{r}^2}$ 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= $\int \mathrm{\rho }\mathrm{d}\mathrm{V}=\int Kr4\pi r^{2}dr$

 .q= 4 $\pi K\frac{R^4}{4}=2e,soK=\frac{2e}{\pi R^4}$

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

F=eE=- $\frac{eK{r}^2}{4{\epsilon }_0}$ but the force applied by proton F= $\frac{e^2}{4\pi {\epsilon }_0({2r)}^2}$

By adding these F= - $\frac{eK{r}^2}{4{\epsilon }_0}+\frac{e^2}{4\pi {\epsilon }_0({2r)}^2}$

If F=0 then $\frac{eK{r}^2}{4{\epsilon }_0}=\frac{e^2}{4\pi {\epsilon }_0({2r)}^2}$  so after solving r= $\frac{R}{8^{1/4}}$

4. 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. 

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

From plate 1 ,E₁= $\frac{-Q}{S\left(2{\epsilon }_0\right)}totheleft$

From plate 2 ,E₂= $\frac{q}{S\left(2{\epsilon }_0\right)}totheright$

Total electric field E= E₁ + E₂ = $\frac{q-Q}{S\left(2{\epsilon }_0\right)}$ to the left , if Q>q

electric field at o due to plate α = $\frac{-Q}{S\left(2{\epsilon }_0\right)}totheleft$

electric field at o due to plate β = $\frac{q1}{S\left(2{\epsilon }_0\right)}totheright$

electric field at o due to plate γ = $\frac{q2}{S\left(2{\epsilon }_0\right)}totheleft$

as the electric field at o is zero therefore

As there is no loss of charge on collision

Q+q= $\frac{Q+q_2}{S\left(2{\epsilon }_0\right)}=\frac{q_1}{S\left(2{\epsilon }_0\right)}$

On solving these

 q₁= (Q+q/2)= charge on plate β

q₂= (q/2)= charge on plate γ