Question

A small object, which has a charge q = 5.6C and mass m = 5.4 x 10-5 kg, is placed in a constant electric field. Starting from rest, the object accelerates to a speed of 4.6 x 103 m/s in a time of 1.0 s. Determine the magnitude of the electric field.

Answer 1

ELECTRIC CHARGE

Atoms and molecules have the same number of protons as electrons and are neutral (without overall charge). Electrons can be transferred from one object to another. When this happens, there is an excess of electrons in one place and a deficiency of electrons in another. Charge is the result of an excess or deficiency of electrons. Where an object has excess electrons, the object is negatively charged. Where there is a deficiency of electrons, the object is positively charged. Electric charge is usually represented in equations by the letter q or Q.

ELECTROSTATIC

"Electrostatic" pertains to electric charges at rest or to fields or phenomena produced by stationary charge(s).

COULOMB'S LAW

Two charges, Q and q, separated by a distance, r, each experience a force of magnitude

F = k|Qq|/r2 where k = 9 x 109 and |Qq| is the positive value of the product of Q and q

Charges of the same sign repel, of different signs attract each other.

ELECTRIC FIELD

The charge , Q, causes an electric force on every other charge, q. Q is called the source charge as it is considered to be the cause of the electric field, while q is called a test charge. The field is a vector quantity. The direction of the field is by definition the direction of the force on a positively charged object: the field points away from a positive source charge, and toward a negative source charge.

A common definition for electric field is a region of space where a positive test charge experiences a force.

Electric field intensity (sometimes just called electric field) is the force per unit charge experienced by a point charge somewhere in space.

El = F/q

Therefore, F = Elq

That is, electric field is the force produced by a source charge, Q, exerted on every coulomb of charge of a test charge at a distance r away from the source of the field.

El = kQ/r2

For parallel plates, El = ?V/d, where ?V is the potential between plates, and d is the distance between the plates.

For parallel plates, El = ?/?0 where ? is the surface charge density C / m2.

GAUSS'S THEOREM

The total flux, ?, through a closed surface is equal to 1/?0 times the total charge contained within it. The location of the charge(es) within the sphere does not matte

You should be able to calculate the acceleration easily from the initial and final velocity, correct?

Next, calculate the force on the charge from Newton's second law, F=ma.

Last, set up your equation for force: F = qE.

E=F/q

F=ma

F=m(v/t)

E=m v/(t q)=

acceleration = speed/time = 4.6 x 10^3

E = F/q = ma/q = 4.44 x 10^-2 N/CNewton / Coulomb

However as given the force on the charge particle is constant (constant electric field), in reality it is not and diminishes as the charge moves with a square of the distance between E producing source.