NCERT : Electric Charges & Fields

Q1. What is the force between two small charged spheres having charges 2 × 10-7 C  and 3 × 10-7 C placed 30cm apart in air ?

Sol. Given q1 = 2 × 10-7 C , q2 = 3 × 10-7 C ,

r = 30 cm = 0.3 m , F = ?

By formula ,\displaystyle F = \frac{1}{4\pi \epsilon_0 }\frac{q_1 q_2}{r^2}

\displaystyle F = (9\times 10^9) \frac{2\times 10^{-7}\times 3\times 10^{-7}}{(0.3)^2}

= 6 × 10-3 N (repulsive)

Q2. The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge −0.8 μC in air is 0.2 N .

(a) What is the distance between the two spheres ?

(b) What is the force on the second sphere due to the first ?

Sol. Given , q1 = 0.4 μC  , q2 = −0.8 μC

F12 = 0.2 N

(a) We know that

\displaystyle F = \frac{1}{4\pi \epsilon_0 }\frac{q_1 q_2}{r^2}

\displaystyle r^2 = \frac{1}{4\pi \epsilon_0 }\frac{q_1 q_2}{F}

\displaystyle r^2 = (9\times 10^9) \frac{0.4\times 10^{-6}\times 0.8\times 10^{-6}}{0.2}

r2 = 144 × 10-4

and r = 12 × 10-2 m

r = 12 cm

(b) Force on second sphere due to first is same i.e. 0.2 N

Q3. Check that the ratio ke2/Gmemp is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify ?

Sol. An electron and proton have a charge of 1.6 × 10-19 C each and their masses are 9.1×10-31 kg and 1.67 ×10-27 kg .

Electrostatic force b/w electron and proton ,

\displaystyle F_e = \frac{1}{4\pi \epsilon_0 }\frac{e .e}{r^2}

Gravitational force b/w electron and proton ,

\displaystyle F_g = G\frac{m_e . m_p}{r^2}

On dividing ,

\displaystyle \frac{F_e}{F_g} = \frac{e^2}{4\pi \epsilon_0 G m_e m_p}

On putting the values ,

\displaystyle \frac{F_e}{F_g} \approx 2.4 \times 10^{39}

The ratio is quite large . This shows that electrostatic forces are much stronger than gravitational forces .

A common example is the lifting of a paper by charged comb against the force of entire Earth on that paper .

Q:4. (a) Explain the meaning of the statement ‘electric charge of a body is quantised ’.
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges ?

Sol. (a) Quantisation of electric charge : It is now well known fact that all charges occurring in nature are positive or negative integral multiple of a basic unit of electric charge which we take as the magnitude of the charge on an electron . Hence charge on an electron is -e and that on a proton happens to be +e , while charge on a neutron is zero . Any charged body will have ±ne  charge , where n is an integer . This fact is called the quantisation of electric charge .

(b) At the macroscopic level one deals with charges that are enormous compared to the magnitude of charge e . Since e = 1.6 × 10-19 C , a charge of magnitude , say 1μC contains something like 1013 times the electronic charge . At this scale , the fact that the charge can increase or decrease only in units of e is not very different from saying that charge can take continuous value . Thus at the macroscopic level , the  quantisation has no practical consequence and can be ignored .

Q:5. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

Sol. Charged is neither created nor destroyed . It is merely transferred from one body to another . Electrons are transferred from glass to silk , so glass has positive charge and silk has negative charge .

Q:6. Four point charges qA = 2 μC, qB = –5 μC, qC = 2 μC, and qD = –5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square ?

Sol. The center O of square is at equal distance of 10/√2 cm from each corners . Since opposite corners have equal charges , forces along both diagonals will be balanced . Resultant force on 1 μC charge at O will be zero .

Q:7. (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
(b) Explain why two field lines never cross each other at any point ?

Sol. (a) They start from a positive charge and end at a negative charge . They are continuous , because force is continuous . They do not have sudden breaks , otherwise a moving test charge will have  to take jumps .

(b) Two lines of force do not intersect each other . If they intersect at a point , there will be two directions of field at that point . Since it is impossible , hence they don’t intersect .

Q: 8. Two point charges qA = 3 μC and qB = -3 μC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 × 10-9 C is placed at this point, what is the force experienced by the test charge ?

Sol: qA = 3 μC = 3 × 10-6 C and qB = -3 μC = -3 × 10-6 C

(a) E0 = ?,

q0 = 1.5 × 10-9 C ,

Electrostatics

E at O due to charge at A,

\displaystyle E_A = \frac{1}{4\pi \epsilon_0}\frac{q_A}{r_A^2}

\displaystyle E_A = (9\times 10^9) \frac{3\times 10^{-6}}{(0.1)^2}

= 27 × 105 N/C along OB

Also, E at O due to charge at B,

EB = 27 × 105 N/C along OB

Resultant E at O ,

EO = EA + EB

= 27 × 105 + 27 × 105

= 54 × 105 N/C along OB

(b) For point O, E = 54 × 105 N/C , q0 = 1.5 × 10-9 C ,

By relation, F = q0 E

F = 1.5 × 10-9 × 54 × 105

= 81 × 10-4

= 8.0 × 10-3 N along OA.

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