plzz give the correct ans.. take your time but ans should b right tree
Answers
Two charges 1 µC and 9 µC are placed at separation of 1 m.
Third charge should be placed in between both charges so that it experience no net force due to these charges
as in that case force with be in opposite direction but equal magnitude
Let say charge q distance from 1 µC is x m
then distance from 9 µC is 1- x m
F = kq₁q₂/r²
k(1)q/x² = k(9)q/(1-x)²
=> (1 - x)² = 9x²
=> x² - 2x + 1 = 9x²
=> 8x² + 2x - 1 =0
=> 8x² + 4x - 2x - 1 = 0
=> 4x(2x + 1) - 1(2x + 1) = 0
=> (2x + 1)(4x - 1) = 0
=> x = -1/2 x = 1/4
-ve value not possible as then forces will be in same direction and will not cancel each otther
x = 1/4
1/4 m from charge 1 µC & 3/4 m from charge 1 µC
please mark this answers brainlist
ok friend I hope this answers is satisfy you
Answer:Given conditions ⇒
q₁ = 12 μC = 12 × 10⁻⁶ C.
q₂ = 8 μC = 8 × 10⁻⁶ C.
Distance between them(r₁) = 10 cm. = 0.1 m.
Final Distance between them(r₂) = 6 cm . = 0.06 m.
Work done in bringing the charges upto the distance of 6 cm is given by,
W = k q₁ q₂ (1/r₂ - 1/r₁)
= 9 × 10⁹ × 12 × 10⁻⁶ × 8 × 10⁻⁶(1/0.06 - 1/0.1)
= 864 × 10⁻³ (16.67 - 10)
= 864 × 10⁻³(6.67)
= 5760 × 10⁻³
= 5.76 J.
Hence, the work done is 5.76 J.
Hope it helps
Explanation:
DO IN THIS PROCESS