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NCERT Exemplar · Class 12 Physics Electric Charges and Fields

31 questions

Short Answer Questions 20–25 (part 4 of 5)

  1. Exercise 20

    A paisa coin is made up of Al-Mg alloy and weighs 0.75g. It has a square shape and its diagonal measures 17\displaystyle 17 mm. It is electrically neutral and contains equal amounts of positive and negative charges. Treating the paisa coins made up of only Al, find the magnitude of equal number of positive and negative charges. What conclusion do you draw from this magnitude?
    NCERT’s answer
    $\displaystyle 1$ Molar mass \(\displaystyle M\) of Al has \(\displaystyle N_A=6.023 \times 10^{23}\) atoms. \(\displaystyle \therefore m=\) mass of Al paisa coin has \(\displaystyle N=N_A \frac{m}{M}\) atoms Now, \(\displaystyle Z_{\mathrm{Al}}=13, M_{\mathrm{Al}}=26.9815 \mathrm{~g}\) Hence \(\displaystyle N=6.02 \times 10^{23}\) atoms \(\displaystyle / \mathrm{mol} \times \frac{0.75}{26.9815 \mathrm{~g} / \mathrm{mol}}\) \[\begin{aligned} & =1.6733 \times 10^{22} \text { atoms } \\ \therefore q \quad & =+ \text { ve charge in paisa }=N \mathrm{Ze} \\ & =\left(1.67 \times 10^{22}\right)(13)\left(1.60 \times 10^{-19} \mathrm{C}\right) \\ & =3.48 \times 10^4 \mathrm{C} . \end{aligned} \] \(\displaystyle q=34.8 \mathrm{kC}\) of \(\displaystyle \pm \mathrm{ve}\) charge. This is an enormous amount of charge. Thus we see that ordinary neutral matter contains enormous amount of ± charges.
    \[N=\frac{m}{M}\,N_A=\frac{0.75\ \mathrm{g}}{27\ \mathrm{g\,mol^{-1}}}\times6.022\times10^{23}\ \mathrm{mol^{-1}}=1.673\times10^{22}\ \text{atoms} \] Each Al atom has \(\displaystyle Z=13\) protons and $\displaystyle 13$ electrons: \[q=N\,Z\,e=1.673\times10^{22}\times13\times1.6\times10^{-19}\ \mathrm{C}=3.48\times10^{4}\ \mathrm{C}=34.8\ \mathrm{kC} \] The coin holds $\displaystyle 34.8$ kC of each sign, yet its net charge is zero because the two cancel exactly. Ordinary matter is packed with charge, and even a tiny imbalance would give enormous forces.Answer: \(\displaystyle q=3.48\times10^{4}\ \mathrm{C}\) of each sign; matter holds huge charge, balanced exactly.

  2. Exercise 21

    Consider a coin of Example 1.20. It is electrically neutral and contains equal amounts of positive and negative charge of magnitude 34.8\displaystyle 34.8 kC. Suppose that these equal charges were concentrated in two point charges seperated by (i) 1\displaystyle 1 cm (∼12×\displaystyle \sim \frac{1}{2} \times diagonal of the one paisa coin), (ii) 100\displaystyle 100 m (~ length of a long building), and (iii) 106 m\displaystyle 10^6 \mathrm{~m} (radius of the earth). Find the force on each such point charge in each of the three cases. What do you conclude from these results?
    NCERT’s answer
    (i)
    \(\displaystyle F_1=\frac{|\mathrm{q}|^2}{4 \pi \varepsilon_0 \mathrm{r}_1^2}=\left(8.99 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2}\right) \frac{\left(3.48 \times 10^4 \mathrm{C}\right)}{10^{-4} \mathrm{~m}^2}=1.1 \times 10^{23} \mathrm{~N}\)
    (ii)
    \(\displaystyle \frac{F_2}{F_1}=\frac{\mathrm{r}_1^2}{\mathrm{r}_2^2}=\frac{\left(10^{-2} \mathrm{~m}\right)^2}{\left(10^2 \mathrm{~m}\right)^2}=10^{-8} \Rightarrow F_2=F_1 \times 10^{-8}=1.1 \times 10^{15} \mathrm{~N}\)
    (iii)
    \(\displaystyle \frac{F_3}{F_1}=\frac{\mathrm{r}_1^2}{\mathrm{r}_3^2}=\frac{\left(10^{-2} \mathrm{~m}\right)^2}{\left(10^6 \mathrm{~m}\right)^2}=10^{-16}\)
    \[F_3=10^{-16} F_1=1.1 \times 10^7 \mathrm{~N} . \]
    Conclusion: When separated as point charges these charges exert an enormous force. It is not easy to disturb electrical neutrality.
    \[q=34.8\ \mathrm{kC}=3.48\times10^{4}\ \mathrm{C} \] \[F=\frac{1}{4\pi\varepsilon_0}\frac{q^{2}}{r^{2}}=\frac{9\times10^{9}\ \mathrm{N\,m^{2}C^{-2}}\times(3.48\times10^{4}\ \mathrm{C})^{2}}{r^{2}}=\frac{1.09\times10^{19}\ \mathrm{N\,m^{2}}}{r^{2}} \] \[\text{(i)}\ r=10^{-2}\ \mathrm{m}:\quad F=\frac{1.09\times10^{19}}{10^{-4}}=1.09\times10^{23}\ \mathrm{N} \] \[\text{(ii)}\ r=100\ \mathrm{m}:\quad F=\frac{1.09\times10^{19}}{10^{4}}=1.09\times10^{15}\ \mathrm{N} \] \[\text{(iii)}\ r=10^{6}\ \mathrm{m}:\quad F=\frac{1.09\times10^{19}}{10^{12}}=1.09\times10^{7}\ \mathrm{N} \] \[m=\frac{F}{g}=\frac{1.09\times10^{7}\ \mathrm{N}}{9.8\ \mathrm{m\,s^{-2}}}\approx1.1\times10^{6}\ \mathrm{kg} \] Each force is attractive, equal and opposite on the two charges. Even an Earth radius apart, the pull equals the weight of a million kg, so such charges can never be pulled apart: bulk matter stays neutral and its free charge is tiny beside $\displaystyle 34.8$ kC.Answer: \(\displaystyle F\approx1.1\times10^{23}\ \mathrm{N},\ 1.1\times10^{15}\ \mathrm{N},\ 1.1\times10^{7}\ \mathrm{N}\) for (i), (ii), (iii).

  3. Exercise 22

    Fig. 1.10\displaystyle 1.10 represents a crystal unit of cesium chloride, CsCl. The cesium atoms, represented by open circles are situated at the corners of a cube of side 0.40nm, whereas a Cl atom is situated at the centre of the cube. The Cs atoms are deficient in one electron while the Cl atom carries an excess electron.
    (i)
    What is the net electric field on the Cl atom due to eight Cs atoms?
    (ii)
    Suppose that the Cs atom at the corner A is missing. What is the net force now on the Cl atom due to seven remaining Cs atoms?
    NCERT_Question_Class12_Physics_Exemplar_Ch1_Q22
    NCERT’s answer
    (i)
    Zero, from symmetry.
    (ii)
    Removing a +ve Cs ion is equivalent to adding singly charged -ve Cs ion at that location.
    Net force then is
    \[F=\frac{e^2}{4 \pi \varepsilon_0 \mathrm{r}^2} \]
    where \(\displaystyle r=\) distance between the Cl ion and a Cs ion.
    \[\begin{aligned} & =\sqrt{(0.20)^2+(0.20)^2+(0.20)^2} \times 10^{-9}=\sqrt{3(0.20)^2} \times 10^{-9} \\ & =0.346 \times 10^{-9} \mathrm{~m} \end{aligned} \]
    Hence,
    \[\begin{aligned} F & =\frac{\left(8.99 \times 10^9\right)\left(1.6 \times 10^{-19}\right)^2}{\left(0.346 \times 10^{-9}\right)^2}=192 \times 10^{-11} \\ & =1.92 \times 10^{-9} \mathrm{~N} \end{aligned} \]
    Ans \(\displaystyle 1.92 \times 10^{-9} \mathrm{~N}\), directed from A to \(\displaystyle \mathrm{Cl}^{-}\)
    \[q_{\mathrm{Cs}}=+e,\quad q_{\mathrm{Cl}}=-e,\quad a=0.40\ \mathrm{nm} \]
    (i)
    Opposite corners lie at equal distances from Cl, on opposite sides, so their fields cancel pair by pair:
    \[\mathbf{E}_{\mathrm{A}}+\mathbf{E}_{\mathrm{A}'}=\mathbf{0}\ \ (\text{and 3 more pairs})\ \Rightarrow\ \mathbf{E}_{\text{net}}=\mathbf{0} \]
    (ii)
    All eight give zero, so the seven give the opposite of A's pull:
    \[\mathbf{F}_{7}=\mathbf{0}-\mathbf{F}_{\mathrm{A}}=-\mathbf{F}_{\mathrm{A}} \]
    Caution: \(\displaystyle r\) is half the body diagonal, not \(\displaystyle a/2\).
    \[r=\frac{\sqrt3}{2}\,a=0.866\times0.40\ \mathrm{nm}=3.46\times10^{-10}\ \mathrm{m} \]
    \[F=\frac{1}{4\pi\varepsilon_0}\frac{e^{2}}{r^{2}}=\frac{9\times10^{9}\times(1.6\times10^{-19})^{2}}{(3.46\times10^{-10})^{2}}=1.9\times10^{-9}\ \mathrm{N} \]
    NCERT_Solution_Class12_Physics_Exemplar_Ch1_Q22
    The unbalanced pull is towards \(\displaystyle \mathrm{A}'\), the corner opposite A.
    Answer: (i) \(\displaystyle \mathbf{E}=\mathbf{0}\); (ii) \(\displaystyle F\approx1.9\times10^{-9}\ \mathrm{N}\), along the body diagonal, away from the vacant corner A.

  4. Exercise 23

    Two charges q\displaystyle q and −3q\displaystyle -3 q are placed fixed on x\displaystyle x-axis separated by distance 'd\displaystyle d'. Where should a third charge 2q\displaystyle 2 q be placed such that it will not experience any force?
    NCERT’s answer
    At P: on 2q, Force due to \(\displaystyle q\) is to the left and that due to \(\displaystyle -3 q\) is to the right. \[\begin{aligned} & \therefore \frac{2 q^2}{4 \pi \varepsilon_0 x^2}=\frac{6 q^2}{4 \pi \varepsilon_0(d+x)^2} \\ & \therefore(d+x)^2=3 x^2 \\ & \therefore 2 x^2-2 d x-d^2=0 \\ & x=\frac{d}{2} \pm \frac{\sqrt{3} d}{2} \end{aligned} \] (–ve sign would be between \(\displaystyle q\) and \(\displaystyle -3 q\) and hence is unaceptable.) \[x=\frac{d}{2}+\frac{\sqrt{3} d}{2}=\frac{d}{2}(1+\sqrt{3}) \text { to the left of } q . \] NCERT_Solution_Class12_Physics_Exemplar_Ch1_Q23_ncert
    Take \(\displaystyle q\) at \(\displaystyle x=0\) and \(\displaystyle -3q\) at \(\displaystyle x=d\). The \(\displaystyle 2q\) charge feels no force where the net field is zero. Between the charges both fields point the same way, and beyond \(\displaystyle -3q\) the larger charge is nearer. So the zero lies outside, on the side of the smaller charge \(\displaystyle q\), at distance \(\displaystyle x\) from it: \[E_{q}=E_{-3q}\ \Rightarrow\ \frac{1}{4\pi\varepsilon_0}\frac{q}{x^{2}}=\frac{1}{4\pi\varepsilon_0}\frac{3q}{(x+d)^{2}} \] \[\frac{x+d}{x}=\sqrt3\ \Rightarrow\ x=\frac{d}{\sqrt3-1}=\frac{\sqrt3+1}{2}\,d\approx1.37\,d \] NCERT_Solution_Class12_Physics_Exemplar_Ch1_Q23 \[\mathbf{F}_{2q}=2q\,\mathbf{E}_{\text{net}}=\mathbf{0} \]Answer: at \(\displaystyle \tfrac{\sqrt3+1}{2}d\approx1.37\,d\) from \(\displaystyle q\), on the line of the charges, on the side away from \(\displaystyle -3q\).

  5. Exercise 24

    Fig. 1.11\displaystyle 1.11 shows the electric field lines around three point charges A, B and C.
    (a)
    Which charges are positive?
    (b)
    Which charge has the largest magnitude? Why?
    (c)
    In which region or regions of the picture could the electric field be zero? Justify your answer.
    (i)
    near A, (ii) near B, (iii) near C, (iv) nowhere.
    NCERT_Question_Class12_Physics_Exemplar_Ch1_Q24
    NCERT’s answer
    (a)
    Charges A and C are positive since lines of force emanate from them.
    (b)
    Charge C has the largest magnitude since maximum number of field lines are associated with it.
    (c)(i)
    near A. There is no neutral point between a positive and a negative charge. A neutral point may exist between two like charges. From the figure we see that a neutral point exists between charges A and C. Also between two like charges the neutral point is closer to the charge with smaller magnitude. Thus, electric field is zero near charge A.
    (a)
    Lines leave A and C, end on B:
    \[\text{lines out}\Rightarrow q_{\mathrm A}>0,\ q_{\mathrm C}>0;\qquad \text{lines in}\Rightarrow q_{\mathrm B}<0 \]
    (b)
    More lines \(\displaystyle \Rightarrow\) larger \(\displaystyle |q|\); C sends out the most:
    \[N_{\mathrm C}>N_{\mathrm A},\ N_{\mathrm B}\ \Rightarrow\ |q_{\mathrm C}|\ \text{is largest} \]
    (c)
    (i) near A. Between like charges A and C the fields oppose, so a neutral point N lies there, nearer the smaller A. Between unlike C and B they add; near B or C its own field wins.
    \[E_{\mathrm A}=E_{\mathrm C}\ \Rightarrow\ \frac{1}{4\pi\varepsilon_0}\frac{q_{\mathrm A}}{x^{2}}=\frac{1}{4\pi\varepsilon_0}\frac{q_{\mathrm C}}{(d-x)^{2}},\quad q_{\mathrm C}>q_{\mathrm A}\ \Rightarrow\ x<d-x \]
    NCERT_Solution_Class12_Physics_Exemplar_Ch1_Q24
    Answer: (a) A and C; (b) C; (c) (i) near A.

  6. Exercise 25

    Five charges, q\displaystyle q each are placed at the corners of a regular pentagon of side 'a\displaystyle a' (Fig. 1.12\displaystyle 1.12).
    (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 corners (say A) is removed?
    (iii)
    What will be the electric field at O if the charge q\displaystyle q at A is replaced by −q\displaystyle -q?
    (b)
    How would your answer to (a) be affected if pentagon is replaced by n-sided regular polygon with charge q\displaystyle q at each of its corners?
    NCERT_Question_Class12_Physics_Exemplar_Ch1_Q25
    NCERT’s answer
    (a)(i)
    zero
    (ii)
    \(\displaystyle \frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2}\) along \(\displaystyle \overrightarrow{\mathrm{OA}}\)
    (iii)
    \(\displaystyle \frac{1}{4 \pi \varepsilon_0} \frac{2 \mathrm{q}}{\mathrm{r}^2}\) along \(\displaystyle \overrightarrow{\mathrm{OA}}\)
    (b)
    same as (a).
    (a)(i)
    Five equal fields, \(\displaystyle 72^\circ\) apart, add to zero:
    \[\mathbf{E}_O=\sum_{i=1}^{5}\mathbf{E}_i=\mathbf{0} \]
    (ii)
    Removing A leaves the opposite of A's field:
    \[\mathbf{E}_O=\mathbf{0}-\mathbf{E}_{\mathrm A}=-\mathbf{E}_{\mathrm A},\qquad r=\frac{a}{2\sin36^\circ}=0.85\,a \]
    \[E_O=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^{2}}=\frac{4\sin^{2}36^\circ}{4\pi\varepsilon_0}\frac{q}{a^{2}}\approx\frac{1.38}{4\pi\varepsilon_0}\frac{q}{a^{2}}\quad\text{(from O towards A)} \]
    (iii)
    The four others give \(\displaystyle -\mathbf{E}_{\mathrm A}\), and \(\displaystyle -q\) at A adds a field in the same direction:
    \[E_O=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^{2}}+\frac{1}{4\pi\varepsilon_0}\frac{q}{r^{2}}=\frac{2}{4\pi\varepsilon_0}\frac{q}{r^{2}}\approx\frac{2.76}{4\pi\varepsilon_0}\frac{q}{a^{2}}\quad\text{(from O towards A)} \]
    NCERT_Solution_Class12_Physics_Exemplar_Ch1_Q25
    (b)
    For any regular \(\displaystyle n\)-gon:
    \[r=\frac{a}{2\sin(\pi/n)} \]
    \[\text{(i) } E_O=0,\qquad \text{(ii) } E_O=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^{2}},\qquad \text{(iii) } E_O=\frac{2}{4\pi\varepsilon_0}\frac{q}{r^{2}} \]
    Cases (ii) and (iii) point from O towards the changed corner.
    Answer: (a) (i) $\displaystyle 0$; (ii) \(\displaystyle \dfrac{q}{4\pi\varepsilon_0 r^{2}}\) towards A; (iii) \(\displaystyle \dfrac{2q}{4\pi\varepsilon_0 r^{2}}\) towards A; (b) the same for any \(\displaystyle n\).