4. [20 points] Four point charges are arranged as shown in the diagram at right.

Question

4. [20 points] Four point charges are arranged as shown in the diagram at right. The four point charges are as follows: o- 2 nC q,--4 nC, q q -3 nC. The-4 nC charge is located 10 cm directly to the left of qo and the two -3 nC charges are also 10 cm away from qo but are at angles of 60 degrees above and below the horizontal 10 cm 0-60° qo qi 0-60 10 cm [5 pts] What is For, the electric force acting on the system charge qo due to charge qu? Give magnitude and direction. Show your work. A. 10 cm 93 B. [5 pts] What is the net electric force on the system charge qo? of an angle from the positive x axis. Show your work. Give magnitude and direction in terms [5 pts] If (for part C only) the system charge (qo) were replaced with a -2 nC charge, what would the net electric force on the new system charge be? Give magnitude and direction. Show your work. C.

Explanation / Answer

a) We know electric force is given as

F=kq1 *q2 /r^2

Thus, F01=kq0*q1/r^2=9*10^9*2*10^{-9}*4*10^{-9}/(.1)^2=72*10^{-7} N towards q1

b)net elcetric force on q0 will be the resultant force on q0 due to all charges present. This can be obtained by resolving components of force in horizontal and vertical direction.

Since q2 and q3 are negative charge and q0 is positive thus the direction of force will be along the line joining then towrds q2 and q3.

Now resolving forces F02 AND F03, by geometry the vertcal components will cancel out as magnitude of verical componets of F02 and F03 wll be same but opposite in direction.

Thus net electric force on q0 will be

F0=k{(q0*q1/r01^2)+(q0*q2 cos 60/r02^2)+(q0*q3 cos 60/r03^2)))}

=9*10^9{(-2*10^{-9}*4*10^{-9}/(.01)^2) +(2*10^{-9}*3*10^{-9}/2*(.01)^2)+(2*10^{-9}*3*10^{-9}/2*(.01)^2)}=

=(9*10^9*2*10^{-9}/(.01)^2))(-4*10^{-9}+1.5*10^{-9}+1.5*10^{-9})= -9*10^9*2*10^{-9}*1*10^-9/(.01)^2))=18*10^{-7} N towards left(towards q1)

c)net elcetric force on q0 will be the resultant force on q0 due to all charges present. This can be obtained by resolving components of force in horizontal and vertical direction.

Since q2 and q3 are negative charge and q0 is also negative thus the direction of force will be along the line joining then towrds q0.

Now resolving forces F02 AND F03, by geometry the vertcal components will cancel out as magnitude of verical componets of F02 and F03 wll be same but opposite in direction.

Thus net electric force on q0 will be

F0=k{(q0*q1/r01^2)-(q0*q2 cos 60/r02^2)-(q0*q3 cos 60/r03^2)))}

=9*10^9{(2*10^{-9}*4*10^{-9}/(.01)^2) -(2*10^{-9}*3*10^{-9}/2*(.01)^2)-(2*10^{-9}*3*10^{-9}/2*(.01)^2)}

=(9*10^9*2*10^{-9}/(.01)^2))(4*10^{-9}-1.5*10^{-9}-1.5*10^{-9})= 9*10^9*2*10^{-9}*1*10^-9/(.01)^2))=18*10^{-7} N towards right

The net cherge magnitude will be same only the direction changes .The direction of net force will be towards right.

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