Example 23.2 Find the Resultant Force Consider three point charges located at th

Question

Example 23.2 Find the Resultant Force

Consider three point charges located at the corners of a right triangle as shown in the figure, where q1 = q3 = 7.30 C, q2 = 2.00 C, and a = 0.130 m. Find the resultant force exerted on q3.

SOLVE IT

Conceptualize Think about the net force on q3. Because charge q3 is near two other charges, it will experience two electric forces. These forces are exerted in different directions as shown in the figure. Based on the forces shown in the figure, estimate the direction of the net force vector.

Categorize Because two forces are exerted on charge q3, we categorize this example as a vector addition problem.

Analyze The directions of the individual forces exerted by q1 and q2 on q3 are shown in the figure. The force

exerted by q2 on q3 is attractive because q2 and q3 have opposite signs. In the coordinate system shown in the figure, the attractive force

is to the left (in the negative x direction).

The force

exerted by q1 on q3 is repulsive because both charges are positive. The repulsive force

makes an angle of 45.0° with the x axis.Use Coulomb's law to find the magnitude of

:

Find the magnitude of the force

:

Find the x and y components of the force

:

Find the components of the resultant force acting on q3:

Express the resultant force acting on q3 in unit-vector form:

Finalize The net force on q3 is upward and toward the left in the figure. If q3 moves in response to the net force, the distances between q3 and the other charges change, so the net force changes. Therefore, q3 can be modeled as a particle under a net force as long as it is recognized that the force exerted on q3 is notconstant.

MASTER ITHINTS: GETTING STARTED | I'M STUCK!

  î +  

N

The force exerted by q1 on q3 is 13. The force exerted by q2 on q3 is 23. The resultant force 3is the vector sum 13 + 23.

Explanation / Answer

The force on q1 = Force due to q2 on q1 + Force due to q3 on q1

= kq1q2/a2 j + kq1q3/2a2 ( -cos 45 i - sin 45 j)

= 9 x 109 x 7.3 x 10-6 x 2 x 10-6/(0.13)2 j+ 9 x 109 x 7.3 x 10-6 x 7.3 x 10-6/2(0.13)2( -cos 45 i - sin 45 j)

= 7.775 j + 14.189 ( -cos 45 i - sin 45 j)

=7.775 j -10.033 i-10.033 j

=-10.033 i -2.258 j

all the best in the course work

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