Back to Directory
NEET PHYSICSEasy

The force between two small charged spheres having charges of 2×10−72 \times 10^{-7} C and 3×10−73 \times 10^{-7} C placed 30 cm apart in the air is:

A

6×10−36 \times 10^{-3} N

B

5×10−35 \times 10^{-3} N

C

4×10−44 \times 10^{-4} N

D

5×10−45 \times 10^{-4} N

Step-by-Step Solution

The force between two point charges in air is given by Coulomb's Law: F=14πε0q1q2r2F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}. Given: k=14πε0=9×109k = \frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 N m2^2 C−2^{-2} q1=2×10−7q_1 = 2 \times 10^{-7} C q2=3×10−7q_2 = 3 \times 10^{-7} C r=30r = 30 cm = 0.30.3 m

Substituting these values: F=(9×109)×(2×10−7)×(3×10−7)(0.3)2F = \frac{(9 \times 10^9) \times (2 \times 10^{-7}) \times (3 \times 10^{-7})}{(0.3)^2} F=54×10−50.09=54×10−59×10−2F = \frac{54 \times 10^{-5}}{0.09} = \frac{54 \times 10^{-5}}{9 \times 10^{-2}} F=6×10−3F = 6 \times 10^{-3} N

(Note: This matches NCERT Physics Class 12, Chapter 1, Exercise 1.1).

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started
Solved: PHYSICS Question for NEET | Sushrut