Get, detailed solutions to the questions of the chapter **Electric Charges and Fields** from NCERT textbooks. The objective is to helping students regarding the pattern of answering the question as per the cbse latest marking scheme.**Cbse.jagranjosh.com** provided you NCERT solutions for classes 12th math and science subjects.

**Get some questions answered here****Q. What is the force between two small charged spheres having charges of 2x10-7 C and 3x10-7 C placed 30 cm apart in air?****Sol.**

Given:

Q1 = 2x10-7 C

Q2 = 3x10-7 C

r = 30x10-2 m

We know,

F=(Q_1 Q_2)/(4πε_o r^2 )

Substituting the given values, we get

F = 6x10-3 N**Q. The electrostatic force on a small sphere of charge 0.4 micro coulomb due to another small sphere of charge -0.8 micro coulomb 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.**

a) From the formula

F=(Q_1 Q_2)/(4πε_o r^2 )

Substituting the values, we get

r = 0.12 m = 12 cm

b) Force on second sphere due to first

F21 = (Q_1 Q_2)/(4πε_o r^2 )

This comes to be -0.2 N, negative sign implying the attractive nature of force. This conforms to Newton’s third law.**Q. 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.**

We know very well

k = 9x109 Nm2/C2

e = 1.6x10-19 C

G = 6.6x10-11 Nm2/kg2

me = 9.11x10-31 kg

mp = 1.6x10-27 kg

Ratio = (Nm^2)/C^2 .C^2.〖kg〗^2/〖Nm〗^2 .1/〖kg〗^2 = dimensionless

And value = 2.4x1039

The ratio signifies that electrical forces are immensely stronger than the gravitational forces.