Consider the following system of three small charged objects as in the diagram s

Question

Consider the following system of three small charged objects as in the diagram shown at left. Qi has a net charge of 3uC of type Nell. The force exerted on Qi is in the negative x direction and has a magnitude of 0.35 N. What can you contend about this situation? (below are some questions to help, follow the assignment structure) 0.4m 0.3m Q, 1. What charge type are Q2 and Qs Explain your reasoning. 2. How does the force on Qi by Q2compare to the force on Qi by Qs? Explain your reasoning. 3. What is the charge on Q2 and Qs? 4. How many electrons would need to be added or removed to result in the charge values for all three charges?

Explanation / Answer

let force on Q1 due to Q2 be F1 and force on Q1 due to Q3 be F2.

components of F1 and F2 along y axis are cancelled out and components along x axis are added.

to satisfy above two conditions, only option can be F1 to be towards Q2 from Q1 and F2 to be towards Q1 from Q3.

so as F1 is attractive in nature, Q2 is of opposite sign to that of Q1 i.e. Q2 is -ve.

similarly as F2 is repulsive, Q3 is of same sign as Q1 and Q3 is +ve.

part 1:

Q2 is -ve.

Q3 is +ve.

part 2:

angle made by F1 with -ve y axis=arctan(0.3/0.4)=36.87 degrees

angle made by F2 with +ve y axis=arctan(0.3/0.4)=36.87 degrees

balancing components along y axis:

F1*cos(36.87)=F2*cos(36.87)

==>F1=F2

hence force on Q1 by Q2 is of same magnitude as force on Q1 by Q3.

part 3:

writing force equation for the system along x axis:

F1*sin(36.87)+F2*sin((36.87)=0.35

using F1=F2

==>F1=0.175 N

==>F2=0.175 N

distance between Q1 and Q2=sqrt(0.3^2+0.4^2)=0.5 m

hence k*Q1*Q2/distance^2=force

where k=coloumb’s constant=9*10^9

==>9*10^9*3*10^(-6)*Q3/0.5^2=0.175

==>Q3=0.175*0.5^2/(9*10^9*3*10^(-6))=1.6204*10^(-6) C

Q2=-1.6204*10^(-6) C

Q4.in order to make same charge in all the charges,

that same charge would be (Q1+Q2+Q3)/3

=1 uC

so charge to be removed from Q1=2 uC

electrons to be removed from Q1=1.25*10^13

charge to be removed from Q3=0.6204*10^(-6)

electrons to be removed from Q3=3.8775*10^12

charge to be added to Q2=2.6204*10^(-6)

electrons to be added=1.6378*10^13

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