Question

Three charges are at the corners of an isosceles triangle as shown in the figure . The $\pm$ 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 ${\rm \mu C}$ charge exerts on the dipole.(Suppose that -axis directs to the top of the image.)

$\theta$ = ____________ ${\rm ^\circ}$ counterclockwise from the +y - direction

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

$\tau$   = _________ N * m

d) Find the direction of the torque.

________clockwise________counterclockwise

Answer 1

Concepts and reason

The concepts required to solve the question are the Coulomb’s force and the torque.

The magnitude of the force exerted on the dipole is calculated using the force between charges located on the two sides of the isosceles triangle.

The direction of the force exerted by the charge on the dipole is calculated using the fact that the electric field is directed from positive to negative.

The torque can be calculated using the expression of torque.

The direction of torque is found by the angle between the force and the perpendicular distance.

Fundamentals

The expression to calculate the force between two charges is,

Here, F is the force between two charges, Q is the charge, q is the charge, and r is the distance between two charges.

The torque is expressed as follows:

Here, r is the perpendicular distance, F is the force, is the torque acting, and is the angle between the vector r and force.

(a)

Calculate the magnitude of the force.

Calculate the magnitude of the force the charge exerts on the dipole.

Calculate the force between charge and .

Substitute for k, for Q , for q , and for r in expression .

.

The force is attractive because the charges are in opposite sign.

Calculate the force between charge and .

Substitute for k, for Q , for q , and for r in expression .

.

The force is repulsive because the charges are in same sign.

Calculate the angle as follows:

The components of the force and are calculated as follows:

The x component of the force is,

Substitute 1125 N for and for in expression .

The y component of the force is,

Substitute 1125 N for and for in expression .

The x component of the force is,

Substitute 1125 N for and for in expression .

The y component of the force is,

Substitute 1125 N for and for in expression .

The magnitude of the net force the charge exerts on the dipole is,

Substitute for and for in expression .

(b)

Calculate the direction of the force charge exerts on the dipole.

The direction of the net force is from the charge to -. It means the direction of the net force is -y.

(c)

The torque is expressed as follows:

Here, r is the perpendicular distance, is the force, is the torque acting, and is the angle between the vector r and force.

Substitute 0.03 m for r , 1125 N for F, and for in expression .

(d)

Find the direction of the torque.

The force is directed along the positive x direction and the force directed along the -x direction.

Thus, the torque has clockwise direction.

Ans: Part a

The magnitude of the force the charge exerts on the dipole is 1688 N.

Part b

The force is in the direction 180⁰ counterclockwise from the +y direction

Part c

The magnitude of the torque exerted on the dipole by the charge is

Part d

The direction of the torque is clockwise.