Three charged particles are placed at the corners of an equilateral triangle of

Question

Three charged particles are placed at the corners of an equilateral triangle of side 1.20 m The charges are Q1 = 7.0 ?C , Q2 = -9.1 ?C , and Q3 = -5.7 ?C .

Part A

Calculate the magnitude of the net force on particle 1 due to the other two.

Express your answers using two significant figures.

Part B

Calculate the magnitude of the net force on particle 2 due to the other two.

Express your answers using two significant figures.

Part C

Calculate the magnitude of the net force on particle 3 due to the other two.

Express your answers using two significant figures.

Explanation / Answer

Q2 and Q3 are along x-axis and Q1 is in first quadrant. Net force on Q1 due to Q2 and Q3:

Let F21: force applied by Q2 on Q1 and F31 is force applied by Q3 on Q1.

Using F21 = kQ1Q2/a2

     = [(8.998 x109 ) x(7 x 10-6 )x(-9.1 x 10-6 )]/(1.2)2 ) N

F21= -0.398 N (negative sign indicates attractive forces)

Similarly, F31 = kQ1Q3/a2

= [(8.998 x109 ) x(-5.7 x 10-6 )x(7 x 10-6 )]/(1.2)2 ) N

= -0.249 N (negative force means attractive forces).

The x and y components of the forces can be calculated as:

F21x = F21 cos60 = -0.398x cos60 = -0.199 N (left)

F21y = F21 cos30 = -0.398x cos30 = -0.344 N (down)

F31x= F31 cos60 = -0.249x cos60 = -0.1245 N (right)

F31y = F31 cos30 = -0.249x cos30 = - 0.215 N (down)

Net Force, Fx = (-0.199) - (-0.1245) N= -0.0745 N (left) and

Fy = (-0.344)+ (-0.215) = -0.559 N (down)

Magnitude of net force, F=(Fx2 +Fy2)

F  =[(-0.0745)2 +(-0.559)2]

=0.5639N

=-0.563 N (negative sign means attractive force)

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