Three charges are at the cornersof an isosceles triangle as shown in the figure ….

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Question “Three charges are at the cornersof an isosceles triangle as shown in the figure ….”

Three charges are at the corners
of an isosceles triangle as shown in the figure . The 5.00 μC charges form a dipole.

a) Find the magnitude of the force the
-10.00 μC charge exerts on the
dipole.
F= _______ N

b)  Find the direction of the force
the -10.00 charge exerts on the
dipole.(Suppose that -axis directs to the top of the image.)

= ____________ counterclockwise from the +y – direction

c) For an axis perpendicular to the line
connecting the 5.00 μC charges at the midpoint of this
line, find the torque (magnitude) exerted on the dipole by the
-10.00 μC charge.

  = _________ N * m

d) Find the direction of the torque.

Answer

To calculate the force between charges, use this expression:

F refers to the force between two charges. Q represents the charge, q the charge and r the distance between the two charges.

Here’s how to express the torque:

Here, r is a perpendicular distance, F the force,

is torque acting and

is angle between vector r and force.

(a)

Calculate the force’s magnitude.

Calculate the magnitude the

charges exerts upon the dipole.

Calculate the force of charge

versus

.

Substitute

to replace k,

for Q,

for q, and

to replace r in expression

.

.

Because the charges are in opposing signs, the force is appealing.

Calculate the force of charge

vs

.

Substitute

to replace k,

for Q,

to replace q, and

to substitute r in expression

.

.

Because the charges are all in one sign, the force is repelling.

The following equations can be used to calculate the angle:

These are the components of force

and

.

The x part of force

,

Substitute 1125N for HTMLmedia_tag_24$, and HTMLmedia_tag_29$ to

in expression

.

The and components of force

are,

Substitute 1125N for HTMLmedia_tag_24$, and HTMLmedia_tag_29$ to

in expression

.

The x part of force

,

Substitute 1125N for HTMLmedia_tag_25$, and HTMLmedia_tag_29$ to

in expression

.

The y component is the force

.

Substitute 1125N for //media_tag_24$, and //media_tag_29$ to

in expression #media_tag_48$.

The net force that the

charges exerts upon the dipole is,

Substitute

to

and

to

within expression

.

(b)

Calculate the direction that the force

charges exerts on the dipole.

The direction of net force is from charge

to –

. This means that the direction of net force is y.

(c)

Here’s how to express the torque:

Here, r is a perpendicular distance.

represents the force.

is the torque acting.

indicates the angle between vector r, force.

Substitute 0.03 M for r, 1125 N to F and

for

for

.

(d)

Determine the direction of the torque.

The force

is directed in the positive -x direction, while the force #media_tag_74$ is directed in the – _x direction.

The torque is therefore in clockwise direction.

Ans: Part A

1688 N is the magnitude of the force that the

charges exerts upon the dipole

Part b

The direction of the force is 180 ⁰ clockwise from the + y direction

Part c

The

charge determines the magnitude of the torque applied to the dipole by the #media_tag_76$ cost.

Part d

The torque direction is clockwise.

Conclusion

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