Practice: P 465 #13, 15 and 16 and  P 489 #1-9, 14-17 13.   The forces on each charge lie along a line connecting the charges.  Let the variable d represent the length of a side of the triangle, and let the variable Q represent the charge at each corner.  Since the triangle is equilateral, each angle is 60o.

The direction of  is in the   y-direction  .  Also notice that it lies along the bisector of the opposite side of the triangle.  Thus the force on the lower left charge is of magnitude , and will point .  Finally, the force on the lower right charge is of magnitude , and will point .

14.   Determine the force on the upper right charge, and then use the symmetry of the configuration to determine the force on the other three charges.  The force at the upper right corner of the square is the vector sum of the forces due to the other three charges.  Let the variable  represent the 0.100 m length of a side of the square, and let the variable  represent the 6.00 mC charge at each corner.

Add the x and y components together to find the total force, noting that .

above the x-direction.

For each charge, the net force will be the magnitude determined above, and will lie along the line from the center of the square out towards the charge.

15.   Determine the force on the upper right charge, and then the symmetry of the configuration says that the force on the lower left charge is the opposite of the force on the upper right charge.  Likewise, determine the force on the lower right charge, and then the symmetry of the configuration says that the force on the upper left charge is the opposite of the force on the lower right charge.

The force at the upper right corner of the square is the vector sum of the forces due to the other three charges.  Let the variable  represent the 0.100 m length of a side of the square, and let the variable  represent the 6.00 mC charge at each corner.

Add the x and y components together to find the total force, noting that .

from the x-direction, or exactly towards the center of the square.

For each charge, the net force will be the magnitude of  and each net force will lie along the line from the charge inwards towards the center of the square.

1.         The work done by the electric field can be found from Eq. 17-2b.

2.         The work done by the electric field can be found from Eq. 17-2b.

3.         The kinetic energy gained is equal to the work done on the electron by the electric field.  The

potential difference must be positive for the electron to gain potential energy.  Use Eq. 17-2b.

4.     The kinetic energy gained by the electron is the work done by the electric force.  Use Eq. 17-2b to calculate the potential difference.

The electron moves from low potential to high potential, so  plate B  is at the higher potential.

5.     The magnitude of the electric field can be found from Eq. 17-4b.

6.  The magnitude of the voltage can be found from Eq. 17-4b.

7.     The distance between the plates can be found from Eq. 17-4b.

8.         The gain of kinetic energy comes from a loss of potential energy due to conservation of energy, and

the magnitude of the potential difference is the energy per unit charge.

The negative sign indicates that the helium nucleus had to go from a higher potential to a lower potential.

9.     Find the distance corresponding to the maximum electric field, using Eq. 17-4b.

14.   Use Eq. 17-5 to find the potential.

15.   Use Eq. 17-5 to find the charge.

16.   The work required is the difference in potential energy between the two locations.  The test charge has potential energy due to each of the other charges, given in Conceptual Example 17-7 as .  So to find the work, calculate the difference in potential energy between the two locations.  Let Q represent the  charge, let q represent the  test charge, and let d represent the 32 cm distance.

An external force needs to do positive work to move the charge.

17.