Question

A charge q1 = 2 nC is placed 10 cm from another charge q2 = -5 nC. A test charge of +3 nC is placed in the middle in between these charges. a) Calculate the magnitude of the force on the test charge. b) What is the direction of this force? Show on a sketch. c) Where can the test charge be placed so that the net force on it is zero? d) Compare the Coulombic force between two electrons separated by the width of a Hydrogen atom to the gravitational force between them.

Answer 1

part a:

force=field*charge

field due to a positive charge is away from the charge while field due to negative charge is towards the charge.

fied at middle between these charges:

field due to q1:

as it is positive, direction is away from q1 and towards q2.

field mangnitude=k*charge/distance^2

=9*10^9*2*10^(-9)/0.05^2

=7200 N/C

field due to q2:

as q2 is negative, it is directed from q1 towards q2.

field magnitude=9*10^9*5*10^(-9)/0.05^2=18000 N/C

so total field=25200 N/C

then force magnitude=field*charge

=7.56*10^(-5) N

part b:

direction of force is from q1 to q2.

part c:

net force will be zero when the test charge is placed to the left of q1.

let distance of test charge from q1 to the left be d.

then distance from q2=d+0.1 meters

balancing the forces due to two charge:

k*q1/d^2=k*q2/(d+0.1)^2

==>2/d^2=5/(d+0.1)^2

==>2*(d^2+0.01+0.2*d)=5*d^2

==>3*d^2-0.4*d-0.02=0

solving for d, we get

d=0.172 m

part d:

width of hydrogen atom=d

electric force/ gravitational force=(k*q1*q2/d^2)/(G*m1*m2/d^2)

=k*q1*q2/(G*m1*m2)

=9*10^9*1.6*10^(-19)*1.6*10^(-19)/(6.674*10^(-11)*9.1*10^(-31)*9.1*10^(-31))

=4.1688*10^42