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NEET PHYSICSEasy

Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5×10−76.5 \times 10^{-7} C? (The radii of A and B are negligible compared to the distance of separation.)

A

1.52×10−21.52 \times 10^{-2} N

B

3.7×10−23.7 \times 10^{-2} N

C

2.01×10−22.01 \times 10^{-2} N

D

2.23×10−12.23 \times 10^{-1} N

Step-by-Step Solution

The force of repulsion between two point charges is given by Coulomb's Law: F=14πε0q1q2r2F = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2}. Given: q1=q2=6.5×10−7q_1 = q_2 = 6.5 \times 10^{-7} C r=50r = 50 cm =0.5= 0.5 m k=9×109k = 9 \times 10^9 N m2^2 C−2^{-2}

Calculation: F=9×109×(6.5×10−7)×(6.5×10−7)(0.5)2F = \frac{9 \times 10^9 \times (6.5 \times 10^{-7}) \times (6.5 \times 10^{-7})}{(0.5)^2} F=9×42.25×10−50.25F = \frac{9 \times 42.25 \times 10^{-5}}{0.25} F=9×169×10−5F = 9 \times 169 \times 10^{-5} F=1521×10−5F = 1521 \times 10^{-5} N F=1.52×10−2F = 1.52 \times 10^{-2} N. (See NCERT Physics Class 12, Exercise 1.12).

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