two charges 1 microcoulomb each of opposite polarity are separated by 0.02 metre the force acting on another one microcoulomb charges are placed at the midpoint of the line joining the two charges will be
Answers
Explanation:
F=kq2÷r2
9×10^9×(1×10^-6)^2÷(0.01)2
9×10^9×1×10^-12÷(1×10^-2)2
9×10^-3÷1×10-4
90 newton
Fnet=2F=2×90=180 newton
The force acting on another 1 micro coulomb charge placed at the midpoint of the line joining the two charges is 45 N.
Given: Two charges 1 micro coulomb each of opposite polarity is separated by 0.02 meter.
To Find: The force acting on another 1 micro coulomb charge placed at the midpoint of the line joining the two charges.
Solution:
- We shall solve the question by applying the formula for the force applied on a point charge due to another point charge at a distance of r.
F = k q1 × q2 / r² [ q1 and q2 are point charges, k = 1/4πε0]
- It is to be noted that like charges repel and unlike charges attract each other.
- According to our question, one positive charge will repel another positive charge and that positive charge will attract the negative charge at the other end. So, we can observe that the net charge will be towards the negative charge if 1 micro coulomb.
Coming to the numerical,
F = k q1 × q2 / r²
Now, q1 = 1 μC, q2 = 1 μC, r = 0.02 m, k = 9 × 10^9 N.m²/C²
So, F = ( 9 × 10^9 × 1 × 10^-6 × 1 × 10^-6 ) / ( 0.02 )²
= ( 9 × 10^-3 ) / (4 × 10^-4 ) N
= 22.5 N
Now, as specified earlier, since the resultant two forces are acting toward the negative charge,
Thus, net force = 2 × F
= 2 × 22.5 N
= 45 N
Hence, the force acting on another 1 micro coulomb charge placed at the midpoint of the line joining the two charges is 45 N.
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