Electric fields and equipotentials

[dasblack]

Homework Statement

The figure below shows a test charge q between the two positive charges. Find the force (in Newtons) on the test charge for q = -4 µC.
a)Give a positive answer if the force is to the right and a negative answer if the force is to the left.

b.)For the previous question, find the electric field (in Newtons/coulomb) at the position of test charge. Again, supply a positive value if the electric field points to the right and a negative value if it points to the left.

Homework Equations

F=ke|q1|q2|/r^2
E=F/qo

The Attempt at a Solution

a)F21=ke(2x10-6)(4x10-6)/(3x10-2)^2 = 80 to the left
F23=ke(4x10-6)(3x10-6)/(2x10-2)^2 = 269.7 to the right
F2 = 190

b)E=190/-4x10-6 = -4.75e7

Homework Statement

How much work (in joules) is done in moving a charge of 2.5μC a distance of 22 cm along an equipotential at 9~V?

Homework Equations

W=-change in PE, where PE = q(change in volts)

The Attempt at a Solution

W=-2.5e-6(9)= -2.25e-5 J

 

[dasblack]

Hmm I'm not sure about the negative for qo

 

[thomate1]

I would like to provide some additional information and clarification about electric fields and equipotentials.

Electric fields are created by electric charges and can be described as the force per unit charge at a given point in space. They are represented by vector arrows that indicate the direction of the force on a positive test charge at that point. The strength of the electric field is determined by the magnitude of the charges and the distance between them.

Equipotentials, on the other hand, are imaginary lines that connect points in space that have the same electric potential. Electric potential is a scalar quantity that represents the amount of work needed to move a unit charge from one point to another. Equipotentials are perpendicular to electric field lines and indicate that no work is done in moving a charge along them.

In the first problem, the force on the test charge can be calculated using Coulomb's law, which states that the force between two charged particles is directly proportional to the magnitude of the charges and inversely proportional to the square of the distance between them. The electric field at the position of the test charge can then be found by dividing the force by the test charge.

In the second problem, the work done in moving a charge along an equipotential can be found by multiplying the charge by the change in electric potential. This is because electric potential is defined as the amount of work needed to move a unit charge between two points.

I hope this additional information helps to clarify the concepts of electric fields and equipotentials.