Answer:
Explanation:
4μC will attract -9μC towards the centre and -5μC will repel it away from the centre. Both these forces are opposite to each other.
Force due to 4μC on -9μC towards the centre
= k x Q₁ Q₂/R² = 9 X 10⁹ X 4 X 10⁻⁶ X 9 X 10⁻⁶ / (1.2)² = 225 X 10⁻³ N/C
Force due to -5μC on -9μC away from the centre
= 9 x 10⁹ x 5 x 10⁻⁶x 9 x 10⁻⁶/( 0.8)² = 632.8 x 10⁻³ .N/C
Ner field =407.8 N/C.
Answer:
0.4078 N towards right
Explanation:
qo = 4 micro coulomb = 4 x 10^-6 C
qA = - 5 micro coulomb = - 5 x 10^-6 C
qB = - 9 micro coulomb = - 9 x 10^-6 Coulomb
OA = 40 cm = 0.4 m
OB = 120 cm = 1.2 m
AB = 120 - 40 = 80 cm = 0.8 m
Force on - 9 micro coulomb due to - 5 micro coulomb is
[tex]F_{AB}=\frac{Kq_{A}q_{B}}{0.8^{2}} =\frac{9\times 10^{9}\times 5\times 10^{-6}\times 9\times 10^{-6}}{0.8^{2}}[/tex]
FAB = 0.6328 N rightwards
Force on - 9 micro coulomb due to 4 micro coulomb is
[tex]F_{OB}=\frac{Kq_{O}q_{B}}{1.2^{2}} =\frac{9\times 10^{9}\times 4\times 10^{-6}\times 9\times 10^{-6}}{1.2^{2}}[/tex]
FOB = 0.225 leftwards
Net force = FAB - FOB = 0.6328 - 0.225 = 0.4078 N towards right
The force acting on 9 micro coulomb so the net force is also along X axis and directed rightwards.
