Question

in the adjoining figure shown three particles A B and C which are equally charged the force acting on B due to a is 2.0 10 to the power minus 6 Newton find out in each figure force exerted on BBC connect one the net force on b ​

Answer 1

Answer:

Part a)

Force on B due to C is given as

$F = 4.5 \times 10^{-6} N$

Part b)

Net force on B due to both charges in case i)

$F = 2.5 \times 10^{-6} N$

Net force on B in case ii)

$F = 4.92 \times 10^{-6} N$

Explanation:

As per Coulomb's law of electrostatic force we know that the force is given as

$F = \frac{kq_1q_2}{r^2}$

$2\times 10^{-6} = \frac{9\times 10^9(q^2)}{(0.015)^2}$

so we have

force exerted by C on B is given as

$F = \frac{9\times 10^9(q^2)}{(0.010)^2}$

from above equations we have

$\frac{F}{2\times 10^{-6}} = \frac{0.015^2}{0.010^2}$

so we have

$F = 4.5 \times 10^{-6} N$

Part b)

Now in first case when all charges are in same line

then net force on the middle charge is given as

$F = F_{bc} - F_{ba}$

$F = 4.5 \times 10^{-6} - 2 \times 10^{-6}$

$F = 2.5 \times 10^{-6} N$

Now in second case when three charges are placed perpendicular

so here we have

$F = \sqrt{F_{ba}^2 + F_{bc}^2}$

$F = \sqrt{4.5^2 + 2^2} \times 10^{-6}$

$F = 4.92 \times 10^{-6} N$

#Learn

Topic : Electrostatic force

https://brainly.in/question/13169785

Answer 2

Answer:

Check this out..Got the answers....If u like it support