## Electrostatics

### Maharashtra Board-Class-11-Science-Physics-Chapter-10

### Solutions

**Question 1. Choose the correct option.**

**(i) A positively charged glass rod is brought close to a metallic rod isolated from ground. The charge on the side of the metallic rod away from the glass rod will be**

**(A) same as that on the glass rod and equal in quantity**

**(B) opposite to that on the glass of and equal in quantity**

**(C) same as that on the glass rod but lesser in quantity**

**(D) same as that on the glass rod but more in quantity**

(C) same as that on the glass rod but lesser in quantity

**(ii) An electron is placed between two parallel plates connected to a battery. If the battery is switched on, the electron will**

**(A) be attracted to the +ve plate**

**(B) be attracted to the -ve plate**

**(C) remain stationary**

**(D) will move parallel to the plates**

(A) be attracted to the +ve plate

**(iii) A charge of +7 ****μ****C is placed at the centre of two concentric spheres with radius 2.0 cm and 4.0 cm respectively. The ratio of the flux through them will be**

**(A) 1:4 **

**(B) 1:2**

**(C) 1:1 **

**(D) 1:16**

(C) 1:1

**(iv) Two charges of 1.0 C each are placed one meter apart in free space. The force between them will be**

**(A) 1.0 N **

**(B) 9 × 10 ^{9} N**

**(C) 9 × 10**^{−}^{9}** N **

**(D) 10 N**

(B) 9 × 10^{9} N

**(v) Two point charges of +5 ****μ****C are so placed that they experience a force of 80****×****10-3 N. They are then moved apart, so that the force is now 2.0****×****10-3 N. The distance between them is now**

**(A) 1/4 the previous distance**

**(B) double the previous distance**

**(C) four times the previous distance**

**(D) half the previous distance**

(B) double the previous distance

**(vi) A metallic sphere A isolated from ground is charged to +50 ****μ****C. This sphere is brought in contact with other isolated metallic sphere B of half the radius of sphere A. The charge on the two sphere will be now in the ratio**

**(A) 1:2 **

**(B) 2:1**

**(C) 4:1 **

**(D) 1:1**

(B) 2:1

**(vii) Which of the following produces uniform electric field?**

**(A) point charge**

**(B) linear charge**

**(C) two parallel plates**

**(D) charge distributed an circular any**

(C) two parallel plates

**(viii) Two point charges of A = +5.0 ****μ****C and B = -5.0 ****μ****C are separated by 5.0 cm. A point charge C = 1.0 ****μ****C is placed at 3.0 cm away from the centre on the perpendicular bisector of the line joining the two point charges. The charge at C will experience a force directed towards**

**(A) point A**

**(B) point B**

**(C) a direction parallel to line AB**

**(D) a direction along the perpendicular bisector.**

(C) a direction parallel to line AB

**Question 2. Answer the following questions.**

**(i) What is the magnitude of charge on an electron?**

The magnitude of charge on an electron = 1.602 × 10^{−}^{19}C.

**(ii) State the law of conservation of charge.**

In any closed or isolated system, the algebraic sum of all the electric charges is constant, i.e., the net charge cannot be created or destroyed.

**(iii) Define a unit charge.**

- The SI unit of charge is the coulomb (C).
- Using Coulomb's law, it is defined as follows : One coulomb is the amount of charge, which placed at a distance of one metre from another charge of the same magnitude in vacuum, experiences a force of 9.0 x 10
^{9}

**(iv) Two parallel plates have a potential difference of 10 V between them. If the plates are 0.5 mm apart, what will be the strength of electric charge.**

Given : V = 10 V, d = 0.5 mm = 5 x 10^{−}^{4} m

The strength of the electric field,

E = \(\frac{V}{d}\) = \(\frac{10}{5×10^{-4}}\) = 2 x 10^{4} V/m = **2 x 10 ^{4} N/C**

**(v) What is uniform electric field?**

A uniform electric field is an electric field that has the same magnitude and the same direction at all points in a given region.

**(vi) If two lines of force intersect of one point. What does it mean?**

If two lines of force intersect at a point, it would mean electric field has two directions at that point. This is impossible.

**(vii) State the units of linear charge density.**

SI unit of linear charge density λ is (C/m).

**(viii) What is the unit of dipole moment?**

The SI unit of electric dipole moment is the coulomb-metre (C.m). The debye (D) is a non SI unit of electric dipole moment.

1 debye = 3.33564 x 10^{—}^{30} coulomb-metre.

**(ix) What is relative permittivity?**

The relative permittivity (ε_{r}) ( ≡ dielectric constant) of a medium is the ratio of the absolute permittivity of the medium (ε) to the absolute permittivity of free space (≡ vacuum (ε_{0})).

Relative permittivity ε_{r} = \(\frac{ε}{ε_0}\)

**Question 3. Solve numerical examples.**

**(i) Two small spheres 18 cm apart have equal negative charges and repel each other with the force of 6 × 10**^{−}^{3}** N. Find the total charge on both spheres.**

**Given **: r = 18 cm = 18 x 10^{−}^{2}m, F = 6 x 10^{−}^{8 }, \(\frac{1}{4πε_0}\) = 9 x 10^{9} Nm^{2}/C^{2},

q_{1} = q_{2} = q(Say)

F = \(\frac{1}{4πε_0}(\frac{q_1q_2}{r^2})\) = \(\frac{1}{4πε_0}(\frac{q^2}{r^2})\)

∴ q^{2} = F (4πε_{0}) r^{2} = \(\frac{(6×10^{-8})(18×10^{-2})}{9×10^9}\)

= 6 x 2 x 18 x 10^{−}^{21} = 21.6 x 10^{−}^{20}

∴ q = \(\sqrt{21.6}\) x 10^{−}^{10} = **−**4.648 x 10^{−}^{10} ….(as the charge is negative)

∴ 2q = **−**9.296 x 10^{−}^{10} C

This gives the total charge on both spheres.

**(ii) A charge +q exerts a force of magnitude -0.2 N on another charge -2q. If they are separated by 25.0 cm, determine the value of q.**

**Given** : F = — 0.2 N, r = 25.0 cm = 0.25 m, q_{1} = +q, q_{2} = **−**2q, \(\frac{1}{4πε_0}\) = 9 x 10^{9} N.m^{2}/C^{2},

F = \(\frac{1}{4πε_0}(\frac{q_1q_2}{r^2})\) = \(\frac{1}{4πε_0}(\frac{-2q^2}{r^2})\)

∴ -2q^{2} = F (4πε_{0}) r^{2} = \(\frac{(-0.2)×(0.25)^2}{9×10^9}\)

∴ q^{2} = \(\frac{(0.1)×(0.625)}{9×10^9}=\frac{62.5}{9}\) x 10^{−}^{12}

∴ q = \(\frac{\sqrt{62.5}}{3}\) x 10^{−6}

= 2.635 x 10^{−}^{6} C** = 2.635 μC**

**(iii) Four charges of +6 × 10**^{−}^{8}** C each are placed at the corners of a square whose sides a are 3 cm each. Calculate the resultant force on each charge and show its direction on a diagram drawn to scale.**

**Given **: q_{1} = q_{2} = q_{3} = q_{4} = q = +6 × 10^{−}^{8 }C, a = 3 cm = 3 x 10^{−}^{2} m, \(\frac{1}{4πε_0}\) = 9 x 10^{9} N.m^{2}/C^{2}

Each charge is repelled by the other three. The charges being equal, the magnitudes, F_{14}, F_{12} , F_{21} , F_{23}, F_{32}, F_{34}, F_{41} and F_{43} are all equal to

\(\frac{1}{4πε_0}(\frac{q^2}{a^2})\) = \(\frac{(9×10^9)(6×10^{-8})^2}{(3×10^{-2})^2}\) = 36 mN

With \(\vec{F}_{14}=-\vec{F}_{41}\), \(\vec{F}_{12}=-\vec{F}_{21}\), \(\vec{F}_{23}=-\vec{F}_{32}\), and \(\vec{F}_{34}=-\vec{F}_{43}\),

Also, F_{13}, F_{31}, F_{24} and F_{42} are equal to

\(\frac{1}{4πε_0}(\frac{q^2}{2a^2})\) = \(\frac{(9×10^9)(6×10^{-8})^2}{2(3×10^{-2})^2}\) = 18 mN

With the choice of axes as shown in Fig. the unit vectors along \(\vec{CA}\), \(\vec{AC}\), \(\vec{DB}\), and \(\vec{BD}\)

are respectively, \(\frac{-\hat{i}-\hat{j}}{\sqrt{2}}\), \(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\), \(\frac{-\hat{i}+\hat{j}}{\sqrt{2}}\) and \(\frac{\hat{i}-\hat{j}}{\sqrt{2}}\)

The net force on q_{1} is

\(\vec{F}_1=\vec{F}_{12}+\vec{F}_{14}+\vec{F}_{13}\)

= (36)(\(-\hat{j}\)) + (36)(\(-\hat{i}\) ) + (18)\((\frac{-\hat{i}-\hat{j}}{\sqrt{2}})\)

= (36 + 9\((\sqrt{2}\))\((-\hat{i}-\hat{j})\) = **(48.73 mN)\((-\hat{i}-\hat{j})\)**

∴ The net force on q_{3} is

\(\vec{F}_3=-\vec{F}_1\) = (48.73 mN)\((\hat{i}+\hat{j})\)

The net force on q_{2} is

\(\vec{F}_2\) = (48.73 mN)\((-\hat{i}+\hat{j})\)

and the net force on q_{4} is

\(\vec{F}_4=-\vec{F}_2\) = (48.73 mN)\((\hat{i}-\hat{j})\)

F_{1} = F_{2} = F_{3} = F_{4} = 48.73\(\sqrt{2}\) mN = **68.92 mN**

**(iv) The electric field in a region is given by \(\vec{E}\)*** ***= 5.0\(\hat{k}\)** **N/C. Calculate the electric flux Through a square of side 10.0 cm in the following cases.**

**(a) the square is along the XY plane**

**(b) The square is along XZ plane**

**(c) The normal to the square makes an angle of 45 ^{0} with the Z axis.**

**Given** : \(\vec{E}\) = 5.0\(\hat{k}\) V/m, a = 10.0 cm = 0.1 m

Flux, ø = \(\vec{E}.\vec{dS}\)

**(i)** The square is along the xy-plane. The corresponding area vector,

\(\vec{dS}\) = a^{2}\(\hat{k}\) = 0.01 \(\hat{k}\) m^{2}

∴ The electric flux through the square

= \(\vec{E}\)·(0.01 \(\hat{k}\)) = (5 \(\hat{k}\) V/m)·(10^{-2} \(\hat{k}\) m^{2})

= **5 x 10 ^{-2} V.m** ….(as \(\hat{k}\).\(\hat{k}\) = 1)

**(ii)** The square is along the xz-plane.

The corresponding area vector,

\(\vec{dS}\) = a^{2} \(\hat{j}\) = (10^{-2 }m^{2})\(\hat{j}\)

∴ The electric flux = (5 \(\hat{k}\) V/m)·(10^{-2} \(\hat{j}\) m^{2}) = 0 …. (as \(\hat{k}\).\(\hat{j}\) = 0)

Alternatively, the square is parallel to so that electric field lines do not pass through the square. Hence, **the electric flux through the square is zero.**

**(iii)** The normal to the square makes an angle θ = 45° with the z-axis.

cos θ = cos 45^{0} = \(\frac{1}{\sqrt{2}}\)

.'. The electric flux through the square

= E S cos θ

= (5 V/m) (10^{-2} m^{2}) x \(\frac{1}{\sqrt{2}}\)

= 5 × 10^{-2} × 0.7071

**= 3.5355 × 10 ^{-2} V/m**

**(v) Three equal charges of 10 × 10**^{−}^{8}** C respectively, each located at the corners of a right triangle whose sides are 15 cm, 20 cm and 25 cm respectively. Find the force exerted on the charge located at the 90****° ****angle.**

**Given**: q_{A} = q_{B} = q_{C} = 10 x 10^{−}^{8} C

Force on B due to A,

\(\vec{F}_{BA}\) = \(\frac{1}{4πε_0}(\frac{q_A.q_B}{r^2_{AB}})\) = \(\frac{(9×10^9)(10×10^{-8})^2}{(20×10^{-2})^2}\)

= 2.25 x 10^{-3} N

Force on B due to C,

\(\vec{F}_{BC}\) = \(\frac{1}{4πε_0}(\frac{q_C.q_B}{r^2_{CB}})\) = \(\frac{(9×10^9)(10×10^{-8})^2}{(15×10^{-2})^2}\)

= 4 x 10^{-3} N

∴ Resultant force on point B,

|FB| = \(\sqrt{F_{BA}^2+F_{Bc}^2+2F_{BA}.F_{BC}\,cos\,90^0}\)

= \(\sqrt{(2.25×10^{-3})^2+(4×10^{-3} )^2}\)

= 4.589 x 10^{−}^{3} N

Force exerted on charge at point B is **4.589 x 10**^{−}^{3}** N**

**(vi) A potential difference of 5000 volt is applied between two parallel plates 5 cm a part a small oil drop having a charge of 9.6 ×10 ^{−}**

^{19}**C falls between the plates. Find (a) electric field intensity between the plates and (b) the force on the oil drop.**

**Given**: V = 5000 volt, d = 5 cm = 5 x 10^{−}^{2} m, q = 9.6 x 10^{−}^{9} c, a. Electric field intensity (E) = ? Force (F) = ?

(a) Electric field intensity between the plates E = \(\frac{V}{d}=\frac{5000}{5×10^{-2}}\) = **1 x 10 ^{5} N/C**

(b) The force on the oil drop F = qE = (9.6 x 10^{−}^{9}) x (1 x 10^{5}) = **9.6 x 10**^{−14}** N**

**(vii) Calculate the electric field due to a charge of ****−****8.0 × 10**^{−}^{8}** C at a distance of 5.0 cm from it.**

Given : q = —8 x 10^{−}^{8} C, r = 5.0 cm = 5 x 10^{−}^{2} m, \(\frac{1}{4πε_0}\) = 9 x 10^{9} N.m^{2}/C^{2},

The magnitude of electric ﬁeld

E = \(\frac{1}{4πε_0}.\frac{q^2}{r^2}\) = \((9×10^9).\frac{(-8×10^{-8})^2}{(5×10^{-2})^2}\) = \(-\frac{72}{25}\) x 10^{5} = **−**2.88 x 10^{5}

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