Question

A 15 nC charge and a -5 nC charges are places 3 cm apart on in the horizontal direction. A -90 nC charge is placed directly above the -5nC charge at a distance of 1 cm. What is the magnitude of the net electrical force on the -90nC charge due to the other two charges?

Answer 1

here,

q1 = 15 * 10^-9 C

q2 = -5 * 10^-9 C

a = 0.03 m

b = 0.01 m

q3 = 90 * 10^-9 C

force due to q1 , f1 = k * q1 * q3 /(a^2 + b^2 )

f1 = 9*10^9 * 15 * 90 * 10^-9 * 10^-9 /( 0.03^2 + 0.01^2)

f1 = 0.01215 N

theta = arctan(b/a)

theta = 18.43 degree

f1 = 0.01215 * ( cos(18.43) i + sin(18.43) j)

f1 = 0.01215 * ( 0.9487 i + 0.316 j)

force due to q2 , f2 = k * q2 * q3 / b^2

f2 = - 9 * 10^9 * 5 * 10^-9 * 15 * 10^-9/0.01^2

f2 = - 0.00675 N j

net force , f = f1 + f2

f = 0.01215 * ( 0.9487 i + 0.316 j) - 0.00675 N j

f = 0.01215 * 0.9487 i - 0.00291 j N

|f| = sqrt(( 0.01215 * 0.9487 )^2 + 0.00291^2 )

|f| = 0.012 N

the magnitude of the net force is 0.012 N