NCERT Solutions for Class 12 Physics Chapter 2 : Exercises Solution
2.1 Two charges $5 \times 10^{-8}C$ and $-3 \times 10^{-8}C$ are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Answer:
Given, two charge particles
$q_1= 5*10^{-8}C$
$q_2= -3*10^{-8}C$
The separation between two charged particle $d=16cm=0.16m$
Now, let's assume the point P between two charged particles where the electric potential is zero is x meter away from $q_1$ and $( 9-x)$ meter away from $q_2$
So,
The potential at point P :
$V_p=\frac{kq_1}{x}+\frac{kq_2}{0.16-x}=0$
$V_p=\frac{k5*10^{-8}}{x}+\frac{k(-3*10^{-8})}{0.16-x}=0$
$\frac{k5*10^{-8}}{x}=-\frac{k(-3*10^{-8})}{0.16-x}$
$5(0.16-x)=3x$
$x=0.1m=10cm$
Hence the point between two charged particles where the electric potential is zero lies 10cm away from $q_1$ and 6 cm away from $q_2$
Now, Let's assume a point Q which is outside the line segment joining two charges and having zero electric potential .let the point Q lie r meter away from $q_2$ and (0.16+r) meter away from $q_1$
So electric potential at point Q = 0
$\frac{kq_1}{0.16+r}+\frac{kq_2}{r}=0$
$\frac{k5*10^{-8}}{0.16+r}+\frac{k(-3*10^{-8})}{r}=0$
$5r=3(0.16+r)$
$r=0.24m=24cm$
Hence the second point where the electric potential is zero is 24cm away from $q_2$ and 40cm away from $q_1$
2.3 (a) Two charges $2\mu C$ and $-2\mu C$ are placed at points A and B 6 cm apart. Identify an equipotential surface of the system.
Answer:
Given, 2 charges with charges $2\mu C$ and $-2\mu C$ .
An equipotential plane is a plane where the electric potential is the same at every point on the plane. Here if we see the plane which is perpendicular to line AB and passes through the midpoint of the line segment joining A and B, we see that at every point the electric potential is zero because the distance of all the points from two charged particles is same. Since the magnitude of charges is the same they cancel out the electric potential by them.
Hence required plane is plane perpendicular to line AB and passing through the midpoint of AB which is 3cm away from both charges.
10. (a) Two charges –q and +q are located at points (0, 0, –a) and (0, 0, a), respectively. What is the electrostatic potential at the points (0, 0, z) and (x, y, 0) ?
Answer:
1)electric potential at point (0,0,z)
distance from $q_1$
$d_1=\sqrt{0^2+0^2+(0-a-z)^2}=a+z$
distance from $q_2$
$d_2=\sqrt{0^2+0^2+(a-z)^2}=a-z$
Now,
Electric potential :
$V=\frac{kq_1}{a+z}+\frac{kq_2}{a-z}=\frac{2kqa}{z^2-a^2}$
2) Since the point,(x,y,0) lies in the normal to the axis of the dipole, the electric potential at this point is zero.
10. (c) Two charges –q and +q are located at points (0, 0, –a) and (0, 0, a), respectively How much work is done in moving a small test charge from the point (5,0,0) to (–7,0,0) along the x-axis? Does the answer change if the path of the test charge between the same points is not along the x-axis?
Answer:
Since point (5,0,0) is equidistance from both charges, they both will cancel out each other potential and hence potential at this point is zero.
Similarly, point (–7,0,0) is also equidistance from both charges. and hence potential at this point is zero.
Since potential at both the point is zero, the work done in moving charge from one point to other is zero. Work done is independent of the path.
14. Obtain the equivalent capacitance of the network in Figure. For a 300 V supply, determine the charge and voltage across each capacitor.
Answer:
Given.
$C_1=100pF$
$C_2=200pF$
$C_3=200pF$
$C_4=100pF$
Now,
Lets first calculate the equivalent capacitance of $C_2\: and \:C_3$
$C_{23}=\frac{C_2C_3}{C_2+C_3}=\frac{200*200}{200+200}=100pF$
Now let's calculate the equivalent of $C_1\:and\:C_{23}$
$C_{1-23}=C_1+C_{23}=100+100=200pF$
Now let's calculate the equivalent of $C_{1-23} \: and \:C_4$
$C_{equivalent}=\frac{C_{1-23}*C_4}{C_{1-23}+C_4}=\frac{100*200}{100+200}=\frac{200}{3}pF$
Now,
The total charge on $C_4$ capacitors:
$Q_4=C_{equivalent}V=\frac{200}{3}*10^{-12}*300=2*10^{-8}C$
So,
$V_4=\frac{Q_4}{C_4}=\frac{2*10^{-8}}{100*10^{-12}}=200V$
The voltage across $C_1$ is given by
$V_{1}=V-V_{4}=300-200=100V$
The charge on $C_1$ is given by
$Q_1=C_1V_1=100*10^{-12}*100=10^{-8}C$
The potential difference across $C_2\:and\:C_3$ is
$V_2=V_3=50V$
Hence Charge on $C_2$
$Q_2=C_2V_2=200*10^{-12}*50=10^{-8}C$
And Charge on $C_3$ :
$Q_3=C_3V_3=200*10^{-12}*50=10^{-8}C$
20. (f) What meaning would you give to the capacitance of a single conductor?
Answer:
There is no meaning in the capacitor with a single plate factually. but we give it meaning by assuming the second plate at infinity. Hence capacitance of a single conductor is the amount of change required to raise the potential of the conductor by one unit amount.
Class 12 Physics NCERT Chapter 2: Higher Order Thinking Skills (HOTS) Questions
Q1. The stored energy of a capacitor charged to 100 V, is 1 J. The capacitance of the capacitor is -
Answer:
The energy stored in a capacitor,
$U=\frac{1}{2} C V^2$.
Rearrange the formula to solve for capacitance
$C=\frac{2 U}{V^2}$.
On Substituting the values $U=1 J$ and $V=100 \mathrm{~V}$ into the formula,
$C=\frac{2 \times 1}{100^2}=\frac{2}{10000}=0.0002 \mathrm{~F}$.
The capacitance of the capacitor is $2 \times 10^{-4} \mathrm{~F}$.
Q.2 A capacitor is filled by two dielectric materials equally in two configurations as shown in the figure. The dielectric constants of materials are $K_1=1.25$ and $K_2=2.25$ and the capacitances in two configurations are $C_1$ and $C_2$ respectively, find out the approximate value of $\ \frac {C_1}{C_2}$ when expressed as an integer.
Answer:
The first arrangement is a parallel combination.
$
C_1^{\prime}=\frac{K_1 \varepsilon_0(A / 2)}{d}=\frac{1}{2} \frac{K_1 \varepsilon_0 A}{d}
$
and $C_2^{\prime}=\frac{1}{2} \frac{K_2 \varepsilon_0 A}{d}$
$
C_1=C_1^{\prime}+C_2^{\prime}=\frac{1}{2} \frac{\left(K_1+K_2\right) \varepsilon_0 A}{d}
$
The second arrangement is a series combination:
$
\begin{aligned}
C_1^{\prime \prime} & =\frac{K_1 \varepsilon_0 A}{d / 2}=\frac{2 K_1 \varepsilon_0 A}{d} \\
C_2^{\prime \prime} & =\frac{K_2 \varepsilon_0 A}{d / 2}=\frac{2 K_2 \varepsilon_0 A}{d} \\
\Rightarrow C_2 & =\frac{C_1^{\prime \prime} C_2^{\prime \prime}}{C_1^{\prime \prime}+C_2^{\prime \prime}}=\frac{K_1 K_2}{K_1+K_2} \frac{2 \varepsilon_0 A}{d} \\
\Rightarrow \frac{C_1}{C_2} & =\frac{\left(K_1+K_2\right)^2}{4 K_1 K_2}=\frac{49}{48}
\end{aligned}
$
The approximate value of $\frac{49}{48}$ when expressed as an integer is 1.
Q.3 Two equal charges $q$ are placed at a distance of 2a and a third charge -2q is placed at the midpoint. The potential energy of the system is -
Answer:
Potential energy -
$\begin{aligned} & U=\frac{1}{4 \pi \epsilon_0} \frac{q(-2 q)}{a}+\frac{1}{4 \pi \epsilon_0} \frac{q(-2 q)}{a}+\frac{1}{4 \pi \epsilon_0} \frac{q q}{2 q} \\ & =\frac{1}{4 \pi \sigma_0} \frac{\mathrm{q}^2}{a}\left[-2-2+\frac{1}{2}\right] \\ & U=\frac{1}{4 \pi \epsilon_0} \frac{\mathrm{q}^2}{\mathrm{a}}\left(-\frac{7}{2}\right)=-\frac{7 \mathrm{q}^2}{8 \pi \epsilon_0 \mathrm{a}}\end{aligned}$
Q.4 Twenty-seven drops of water of the same size are equally and similarly changed. They are then united to form a bigger drop. By what factor will the electrical potential change?
A) 9 times
B) 27 times
C) 6 times
D) 3 times
Answer:
Let $V_B$ is the potential of a bigger drop and $V_S$ potential of a small drop.
$
\text { use, } V_B=n^{2 / 3} V_s \quad-(1)
$
but $n=27$. put in (1) we get-
$
\begin{aligned}
& V_B=(27)^{2 / 3} V_s \\
& V_B=\left(3^3\right)^{2 / 3} V_s=3^2 V_s \\
& V_B=9 V_S \rightarrow 9 \text { times }
\end{aligned}
$
Hence, the answer is the option 1A
Q.5 Find the total charge (in $\mu C$) stored in the network of capacitors connected between A and B as shown in figure :
Answer:
The given circuit is a balanced bridge. Total capacitance
$\begin{aligned} & C=\frac{2 \times 4}{2+4}+\frac{6 \times 3}{6+3} \\ & =\frac{4}{3}+2=\frac{10}{3} \mu F \\ & Q=C V=\frac{10}{3} \mu F \times 3 V=10 \mu C\end{aligned}$
Hence, the answer is 10.
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