Physics, charged triangle.   Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equilateral triangle, as the drawing shows. The magnitude of each of the charges is 3.92 μC, and the lengths of the sides of a triangle are 5.56 cm. Calculate the magnitude of the net force that each charge experiences. 

Expert Answers

An illustration of the letter 'A' in a speech bubbles

The figure is below. All the forces represented with thin lines are equal in module. The electric field lines exits the positive charges and enters the negative charge. Thus the direction of forces are (see the figure)

`F_(12) =F_(21)= k_e*q^2/R^2=9*10^9*3.92^2*10^-12/0.0556= 2.4874 N` 

away from charges 1 and 2, on the...

Unlock
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Start your 48-Hour Free Trial

The figure is below. All the forces represented with thin lines are equal in module. The electric field lines exits the positive charges and enters the negative charge. Thus the direction of forces are (see the figure)

`F_(12) =F_(21)= k_e*q^2/R^2=9*10^9*3.92^2*10^-12/0.0556= 2.4874 N` 

away from charges 1 and 2, on the 12 line

`F_(13)=F_(23) = k_e*q^2/R^2 =2.4874 N` ,

away from charges 1 and 2 towards 3

`F_31 =F_32=k_e*q^2/R^2 =2.4874 N` ,

away from charge 3 towards 1 and 2

Now the resultant forces on charges 1, 2 and 3 are (see the law of cosine in a triangle)

`F_1 = sqrt(F_12^2+F_13^2-2*F_12*F_13*cos(60)) =2.4874 N`

`F2 =F1 =2.4874 N` by symmetry arguments

`F_3 =sqrt(F_(31)^2 +F_(32)^2-2*F_31*F_32*cos(120)) =4.3083 N`

`F_1` and `F_2` make an angle of 60 degree with the horizontal and `F3` is downwards.

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
Approved by eNotes Editorial Team