1.

3 equal charges which has magnitude of 2×10^6 coulomb is placed at the corners of right angled triangle of sides 3, 4, 5find the force on the charge at right angle corner.

Answer»

See the diagram. 
the force acting on charge at B due to charge at A is
F_1= \frac{1}{4 \pi \epsilon_0} \frac{q_1q_2}{r^2} =9*10^9* \frac{(2*10^6)^2}{4^2}=2.25*10^{21}N
It ACTS in the DIRECTION from A to B.

the force acting on charge at B due to charge at C is
F_2= \frac{1}{4 \pi \epsilon_0} \frac{q_1q_2}{r^2} =9*10^9* \frac{(2*10^6)^2}{3^2}=4*10^{21}N
It acts in the direction from C to B.

Since F₁ and F₂ are at right angles to each other, the RESULTANT F is given by

F= \sqrt{F_1^2+F_2^2}\\F = \sqrt{(2.25*10^{21})^2+(4*10^{21})^2} \\F= \sqrt{21.0625*10^{42}} \\F=\boxed{4.59*10^{21}\ N}


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