Four identical charged particles (q = +10.1 µC) are located on the corners of a

Question

Four identical charged particles (q = +10.1 µC) are located on the corners of a rectangle as shown in the figure below. The dimensions of the rectangle are L = 60.6 cm and W = 16.2 cm.

(a) Calculate the magnitude of the total electric force exerted on the charge at the lower left corner by the other three charges.

answer in N

(b) Calculate the direction of the total electric force exerted on the charge at the lower left corner by the other three charges.

answer is ° (counterclockwise from the +x-axis)

Explanation / Answer

let lower left charge is at origin.

coordinates of all the charges:

lower left =(0,0)

lower right=(L,0)=(0.606,0) m

upper left=(0,W)=(0,0.162) m

upper right=(L,W)=(0.606,0.162) m

force on the lower left charge by lower right charge:

as both the charges are positive, force is repulsive in nature.

direction of force=(0,0)-(L,0)=(-L,0)

distance=sqrt(L^2+0^2)=L

unit vector along the direction of the force=(-L,0)/L=(-1,0)

force magnitude=k*q*q/L^2

where k=coloumb's constant=9*10^9

hence force magnitude=9*10^9*10.1*10^(-6)*10.1*10^(-6)/0.606^2=2.5 N

in vector notation, force =F1=2.5*(-1,0) N

force on the lower left charge by upper left charge:

as both the charges are positive, force is repulsive in nature.

direction of force=(0,0)-(0,W)=(0,-W)

distance=sqrt(W^2+0^2)=W

unit vector along the direction of the force=(0,-W)/W=(0,-1)

force magnitude=k*q*q/W^2

where k=coloumb's constant=9*10^9

hence force magnitude=9*10^9*10.1*10^(-6)*10.1*10^(-6)/0.162^2=34.983 N

in vector notation, force =F2=34.983*(0,-1) N

force on the lower left charge by upper right charge:

as both the charges are positive, force is repulsive in nature.

direction of force=(0,0)-(L,W)=(-L,-W)

distance=sqrt(L^2+W^2)=0.6273 m

unit vector along the direction of the force=(-L,-W)/0.6273=(-0.606,-0.162)/0.6273=(-0.966,-0.2582)

force magnitude=k*q*q/0.6273^2

where k=coloumb's constant=9*10^9

hence force magnitude=9*10^9*10.1*10^(-6)*10.1*10^(-6)/0.6273^2=2.333 N

in vector notation, force =F3=2.333*(-0.966,-0.2582) N

hence total force on the lower left charge=F1+F2+F3=2.5*(-1,0)+34.983*(0,-1)+2.333*(-0.966,-0.2582)=(-4.7537,-35.5853) N

part a:

magnitude of force=sqrt(4.7537^2+35.5853^2)=35.901 N

part b:

the force is in third quadrant.
angle with -ve x axis=arctan(35.5853/4.7537)=82.391 degrees

then in counterclockwise direction, angle with +ve x axis=180+82.391=262.391 degrees

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