Solutions for Exercises of Electrostatics
Q l. Choose the correct option
i) A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential, capacitance respectively are
(A) Constant, decreases, decreases
(B) Increases, decreases, decreases
(C) Constant, decreases, increases
(D) Constant, increases, decreases
ii) A slab of material of dielectric constant k has the same area A as the plates of a parallel plate capacitor and has thickness (3/4d), where d is the separation of the plates. The change in capacitance when the slab is inserted between the plates is
(A) C= $latex \frac{A\varepsilon_{\mathit\circ}}d(\frac{k+3}{4k})$
(B) C= $latex \frac{A\varepsilon_{\mathit\circ}}d(\frac{2k}{k+3})$
(C) C= $latex \frac{A\varepsilon_{\mathit\circ}}d(\frac{k+3}{2k})$
(D) C= $latex \frac{A\varepsilon_{\mathit\circ}}d(\frac{4k}{k+3})$
iii) Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio.
(A) 1:1
(B) 1:2
(C) 2:1
(D) 1:3
iv) Charge +q and -q are placed at points A and B respectively which are distance 2L apart. C is the mid point of A and B. The work done in moving a charge +Q iv)
along the semicircle CRD as shown in the figure below is
(A) $latex \frac{-qQ}{6\pi\varepsilon_{\mathit\circ}L}$
(B) $latex \frac{qQ}{2\pi\varepsilon_{\mathit\circ}L}$
(C) $latex \frac{qQ}{6\pi\varepsilon_{\mathit\circ}L}$
(D) $latex \frac{-qQ}{4\pi\varepsilon_{\mathit\circ}L}$
v) A parallel plate capacitor has circular plates of radius 8 cm and plate separation lmm. What will be the charge on the plates if a potential difference of 100 V is applied?
(A) 1.78 x 10-8C
(B) 1.78 x 10-5C
(C) 4.3 x 104C
(D) 2 x 10-9C
Q 2. Answer in brief.
i) A charge q is moved from a point A above a dipole of dipole moment p to a point B below the dipole in equitorial plane without acceleration. Find the work done in this process.
ii) If the difference between the radii of the two spheres of a spherical capacitor is increased, state whether the capacitance will increase or decrease.
iii) A metal plate is introduced between the plates of a charged parallel plate capacitor. What is its effect on the capacitance of the capacitor?
iv) The safest way to protect yourself from lightening is to be inside a car. Justify.
v) A spherical shell of radius b with charge Q is expanded to a radius a. Find the work done by the electrical forces in the process.
3. A dipole with its charges, -q and +q located at the points (0, -b, 0) and (0 +b, 0) is present in a uniform electric field E. The equipotential surfaces of this field are planes parallel to the YZ planes. (a) What is the direction of the electric field E? (b) How much torque would the dipole experience in this field?
4. Three charges –q, +Q and –q are placed at equal distance on straight line. If the potential energy of the system of the three charges is zero, then what is the ratio of Q: q?
5. A capacitor has some dielectric between its plates and the capacitor is connected to a DC source. The battery is now disconnected and then the dielectric is State whether the capacitance, the energy stored in it, the electric field, charge stored and voltage will increase, decrease or remain constant.
6. Find the ratio of the potential differences that must be applied across the parallel and series combination of two capacitors C1 and C2 with their capacitances in the ratio 1:2, so that the energy stored in these two cases becomes the same.
7. Two charges of magnitudes -4Q and +2 Q are located at points (2a, 0) and (5a, 0) respectively. What is the electric flux due to these charges through a sphere of radius 4a with its centre at the origin?
8. A 6 pF capacitor is charged by a 300 V supply. It is then disconnected from the supply and is connected to another uncharged 3pF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?
C1 = 6 µF = 6 × 10-6 F, V1 = 300 V,
After fully charged connected C1 connected with C2
C2 = 3 µF = 3 × 10-6 F, Energy Loss = ?
Energy stored by the charged capacitor,
$latex E_1=\frac12C_1V_1^2=\frac12\times6\times10^{-6}\times\times300^2$
$latex \therefore E_1=0.27\;J$
When capacitor is disconnected from the battery and connected to another capacitor, both the capacitors are in parallel hence,
Ceq = C1 + C2 = 9μF = 9 × 10-12F
Also the common potential after connection will be,
$latex V_2=V_c=\frac{C_1V_1+C_2V_2}{C_1+C_1}$
Here, V1 is 250 V and V2 is 0, as it was uncharged.
$latex \therefore V_c=\frac{6\times10^{-6}\times300+0}{9\times10^{-6}}=200V$
So, final energy stored by the system is,
$latex E_2\;=\;\frac12C_{eq}V_c^2$
$latex E_2\;=\;\frac12\;\times\;9\;\times\;10^{-6}\;\times\;\left(200\right)^2$
$latex \therefore\;E_2=0.18\;J\;$
Hence energy loss is given as,
$latex \triangle E=E_1-E_2$
$latex \therefore\triangle E\;=\;(0.27\;-\;0.18)$
$latex \therefore\triangle E\;=\;0.09J\;=\;9\;\times\;10^{-2}J$
9. One hundred twenty five small liquid drops, each carrying a charge of 5 pC and each of diameter 0.1 m form a bigger drop. Calculate the potential at the surface of the bigger drop.
q = 0.5 µC = 0.5 × 10-6C,
$latex r=\frac d2=\frac{0.1}2=0.05\;m$
n = 125, V (of bigger drop) = ?
When smaller drops from big drop, volume is same.
$latex n\;\times\;\frac43\pi r^3\;=\;\frac43\pi R^3$
$latex \therefore R^3=\pi r^3$
$latex \therefore\;R\;=\;n^{1/3}r$
$latex \therefore\;R\;=\;\left(25\right)^{1/3}\left(0.05\right)\;=\;0.25\;m$
Total charge on big drop is, Q = nq
$latex \therefore\;Q\;=\;125\;\times\;0.5\;\times\;10^{-6}$
$latex \therefore\;Q\;=\;62.5\;\times\;10^{-6}C\;$
Capacitance of the big drop,
$latex C\;=\;4\pi\varepsilon_0R$
$latex C\;=\;\frac1{9\times10^9}\;\times\;0.25$
$latex C\;=\;0.0277\;\times\;10^{-9}F$
Hence potential of big drop is
$latex V\;=\;\frac{Q\;}C$
$latex V\;=\;\frac{62.5\;\times\;10^{-6}}{0.0277\;\times\;10^{-9}}$
$latex V\;=\;2.25\;\times\;10^6V$
10. The dipole moment of a water molecule is 6.3 x 10–” Cm. A sample of water contains 1021 molecules, whose dipole moments are all oriented in an electric field of strength 2.5 x 105N /C. Calculate the work to be done to rotate the dipoles from their initial orientation θ1= 0 to one in which all the dipoles are perpendicular to the field, θ2 = 90°.
p = 6.3 × 10-30Cm, E = 2.5 ×105V/m,
θ1 = 0,
θ2 = 90°, N = 1021, W = ?
$latex W\;=\;pE(\cos\theta_1\;-\;\cos\theta_2)\;$
∴ W = 6.3 × 10-30 × 2.5 × 105 × (cos0 − cos90°)
∴ W = 6.3 × 10-30 × 2.5 × 105 × (1 − 0)
∴ W = 15.75 × 10-25J
Hence total work for all molecules,
Wtotal = W × N
∴ Wtotal = 15.75 × 10-2 × 1021
= 1.575 × 10-3J
11. A charge 6 pC is placed at the origin and another charge —5 pC is placed on the y axis at a position A (0, 6.0) m.
(a) Calculate the total electric potential at the point P whose coordinates are (8.0, 0) m
(b) Calculate the work done to bring a proton from infinity to the point P ? What is the significance of the negative sign?
q1 = 6μC = 6 × 10-6C, at origin O (0, 0)
q2 = −5μC = −5 × 10-6C, at A (0, 6)
a) VP =? P (8, 0) (b) Work done for bringing proton from infinity to P
$latex a)\;\;V_p\;=\;\frac1{4\pi\varepsilon_0}\frac{q_1}{OP}+\frac1{4\pi\varepsilon_0}\frac{q^2}{AP}$
OP = 8m,
$latex AP\;=\;\sqrt{{(0\;-\;8)}^2\;+\;{(6\;-\;0)}^2}……..by\;dis\tan ce\;formula$
∴ AP = 10m
$latex \therefore\;V_p\;=\;9\;\times\;10^9=\left[\frac{6\times10^{-6}}8+\frac{-5\times10^{-6}}{10}\right]$
$latex \therefore V_p=\;9\;\times\;10^3\;\left[\frac34-\frac12\right]=2.25\times10^3\;V$
b) According to the definition, the potential at a point is defined as the work done per unit charge while moving that charge from infinity to the point.
$latex \therefore V_p=\frac Wq\;\;\;\;\;\;\;\;\therefore W=qv$
$latex \therefore\;W\;=\;qV\;=\;(1.6\;\times\;10^{-19})\;\times\;2.25\times\;10^3$
$latex \therefore\;W\;=\;3.6\;\times\;10J\;\;$
12. In a parallel plate capacitor with air between the plates, each plate has an area of 6 x 10-3 m2 and the separation between the plates is 2 mm. a) Calculate the capacitance of the capacitor, b) If this capacitor is connected to 100 V supply, what would be the charge on each plate? (c) How would charge on the plates be affected if a 2 mm thick mica sheet of k = 6 is inserted between the plates while the voltage supply remains connected?
A = 6 x 10-3 m2, d = 2mm = 2 × 10-3m, For air k =1
$latex a)\;C\;=\frac{A\varepsilon_0k}d$
$latex C\;==\frac{6\times10^{-3}\times8.85\times10^{-12}\times1}{2\times10^{-3}}$
$latex =\;2.655\;\times\;10^{-9}\;C$
b) V = 100 V, Q = ?
$latex C\;=\frac QV\;\;\;\;\;\;\;\therefore Q=CV$
$latex \therefore\;Q\;=\;CV\;=\;2.655\;\times\;10^{-11}\;\times\;100$
$latex =2.655\;\times10^{-9}\;C$
c) Mica sheet (2mm) of k = 6 is inserted. Distance between the plates of the capacitors is also 2 mm. Hence complete is replaced by mica.
$latex \therefore\;C\;=\frac{A\varepsilon_0k}d\;=\;C_{air}\;\times\;k$
$latex \therefore\;C\;=\;2.655\;\times\;10^{-11}\;\times\;6$
$latex \therefore\;C_{new}\;=\;15.93\;\times\;10^{-11}F$
Hence capacitance is increased.But C=Q/V and as potential is kept constant charge on plates increase.
$latex \therefore\;Q_{new}=15.93\times10^{-11}\times100\;$
$latex =15.93\times10^{-9}\;C$
13. Find the equivalent capacitance between P and Q. Given, area of each plate = A and separation between plates = d.
1) The space between the plates of the capacitor is filled with 3 different dielectrics keeping A same. Hence total capacitance is considered as series combination of 3 capacitances each formed by A and d/3.
$latex C_1=\frac{A\varepsilon_0k_1}{d/3},\;C_2=\frac{A\varepsilon_0k_2}{d/3}\;and\;C_3=\frac{A\varepsilon_0k_3}{d/3}$
$latex \therefore\frac1C=\frac1{C_1}+\frac1{C_2}+\frac1{C_3}….\;series\;combination$
$latex \therefore\frac1C=\frac{d/3}{A\varepsilon_0k_1}+\frac{d/3}{A\varepsilon_0k_2}+\frac{d/3}{A\varepsilon_0k_3}$
$latex\therefore\;\frac d{3A\varepsilon_0}\left[\frac32\right]$
$latex \therefore\;C\;=\;\frac{2A\varepsilon_0}d$
2) Here three capacitances are as follows,
$latex C_1\;=\;\frac{\left(A/2\right)\varepsilon_0k_1}{d/3},$
$latex C_2\;=\;\frac{\left(A/2\right)\varepsilon_0k_2}{d/3}$
$latex and\;C_3=\frac{\left(A/2\right)\varepsilon_0k_3}{d/3}$
And the equivalent diagram for C as follows :
Hence C2 and C3 are in series.
$latex \therefore\;\frac1{C_{23}}\;=\;\frac1{C_2}\;+\;\frac1{C_3}$
$latex \therefore\;\frac1{C_{23}}\;=\;\frac{d/2}{\left(A/2\right)\varepsilon_0k_2}\;+\;\frac{d/2}{\left(A/2\right)\varepsilon_0k_3}$
$latex \therefore\;\frac1{C_{23}}=\frac d{A\varepsilon_0}\left[\frac13+\frac16\right]$
$latex \therefore\;\frac1{C_{23}}=\;\frac d{A\varepsilon_0}\;\times\;\frac12$
$latex \therefore\;C_{23}\;=\;\frac{2A\varepsilon_0}d$
Now C23 and C1 are in parallel
$latex \therefore\;C\;=\;C_{23}\;+\;C_1$
$latex \therefore\;C\;=\;\frac{2A\varepsilon_0}d\;+\;\frac{\left(A/2\right)\varepsilon_0\times4}d$
$latex =\frac{A\varepsilon_0}d=\left[2+2\right]$
$latex \therefore C=\frac{4A\varepsilon_0}d$