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:

Post the figure also..

Answer 2

Given:  A B and C are equally charged, force acting on B due to A is 2 x 10^-6.

To find: Force on B due to C, Net force on B due to both charges

Solution:

• Now we have given three particles of equal charges.

• Now Coulomb's law of electrostatic force is :

               F = kq1q2 / r^2

• Now putting values in formula we get:

              2 x 10^-6 = 9 x 10^9 x q^2/ (0.015)^2

• Now force on B due to C will be:

              F = 9 x 10^9 x q^2/ (0.01)^2

• Now equation both equations, we get:

              F / 2 x 10^-6 =  (0.015)^2 /  (0.01)^2

• So, F = 4.5 x 10^-6 N

• Now when all charges are in line, the net force on middle charge will be:

              F = F(bc) - F(ba)

              F = 4.5 x 10^-6  -  2 x 10^-6

              F = 2.5 x 10^-6 N

• But when charges are placed perpendicular, then:

              F = √ ( F(bc)^2 + F(ba)^2 )

              F = √ ( 4.5² + 2²)  x 10^-6

              F = 4.92 x 10^-6 N

Answer:

              So Force on B due to C is 4.5 x 10^-6 N and Net force on B due to both charges in line is 2.5 x 10^-6 N and when perpendicular is 4.92 x 10^-6 N.