Question

1. Field and force with three charges: At a particular moment, three small charged balls, one negative and two positive, are located as shown in (al) What is the clectric field at the location of Qi, due to (a2) Repeat (al), but first calculate the electric potential at the location of Qi, due to O2, and then use that to calculate the electric field at the location of oi, due to 02. (a2) What is the electric field at the location of O, due to os?

Answer 1

part a1:

as Q2 is positive,

field due to Q2 will be directed away from Q2.

so at the location of Q1, field is towards +ve y axis

distance =4 cm

then field magnitude=9*10^9*Q2/distance^2

=9*10^9*7*10^(-9)/0.04^2=39375 N/C

part a2:

electric potential at Q1 due to Q2

=9*10^9*Q2/distance

=9*10^9*7*10^(-9)/0.04=1575 Volts

then electric field=electric potential/distance

=1575/0.04=39375 N/C

part a2:

distance=sqrt(4^2+3^2)=5 cm=0.05 m

field magnitude=9*10^9*Q3/distance^2

=9*10^9*7*10^(-9)/0.05^2=2.52*10^4 N/C

field is directed towards Q3 as Q3 is negative.

vector along this direction=(3,0)-(0,4)=(3,-4)

unit vector=(3,-4)/5=(0.6,-0.8)

part a3:

net field at Q1=39375*(0,1)+2.52*10^4*(0.6,-0.8)

=(1.512*10^4, 19215) N/C

so net field magnitude=sqrt((1.512*10^4)^2+19215^2)=2.4451*10^4 N/C

direction=arctan(19215/(1.512*10^4))=51.8013 degrees , counter clockwise with +ve x axis

part b:

force on Q1 due to Q2 and Q3=electric field*charge

=2.4451*10^4*1*10^(-9)=2.4451*10^(-5) N

as Q1 is positive, force direction will be same as electric field direction.

51.8013 degrees , counter clockwise with +ve x axis