Two identical small balls having charges q1
+1mC and q2
= – 0.33 mC are brought in
contact and them moved apart to a distance r
= 20 cm. Find the force of their interaction.
(a) 50 kN (b) 25 kN
(c) 20 kN (d) 10 kN
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Explanation:
q1 = 1mC = 10^-3 C
q2 = -0.33 mC = -0.33 × 10^-3 C
When brought in contact the charges get distributed equally.
New charge q1 = q2 = q = 1-0.33/2 × 10^-3 = 0.67 × 10^-3 C
Force between charges = F = kq1q2/r^2 = kq^2/r^2 = (9×10^9 × 0.67 × 0.67 × 10^-6)/(20 × 20 × 10^-4 )
F = (4.04 × 10^3)/(400×10^-4) = (404/400)×10^5 = 1.01 × 10^5 = 10×10^4 = 100kN
I think either my calculations might be wrong or the option d is wrong it is either 100 or 10 kN
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