Electric Field?

Question

An electric field of 332,000 N/C points due west at a certain point. What are the magnitude and direction of the force that acts on a charge of -8.9 µC at this spot?
magnitude

Answer 1

The field is described aptly here:

"Electric field strength is a vector quantity; it has both magnitude and direction. The magnitude of the electric field strength is defined in terms of how it is measured. Let's suppose that an electric charge can be denoted by the symbol Q. This electric charge creates an electric field; since Q is the source of the electric field, we will refer to it as the source charge. The strength of the source charge's electric field could be measured by any other charge placed somewhere in its surroundings. The charge that is used to measure the electric field strength is referred to as a test charge since it is used to test the field strength. The test charge has a quantity of charge denoted by the symbol q. When placed within the electric field, the test charge will experience an electric force - either attractive or repulsive. As is usually the case, this force will be denoted by the symbol F. The magnitude of the electric field is simply defined as the force per charge of the test charge.

If the electric field strength is denoted by the symbol E, then the equation can be rewritten in symbolic form as
.

The standard metric units on electric field strength arise from its definition. Since electric field is defined as a force per charge, its units would be force units divided by charge units. In this case, the standard metric units are Newton/Coulomb or N/C.

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From this, we can establish the direction and magnitude of the force (in Newtons) because we know this:

"A stationary point electric charge $q$ is known to produce a scalar potential
\begin{displaymath} \phi = \frac{q}{4 \pi \epsilon_0 r} \end{displaymath} (17.1)

a distance $r$ from the charge. The constant $\epsilon_0 = 8.85 \times 10^{-12} \mbox{ C}^2 \mbox{ N}^{-1} \mbox{ m}^{-2}$ is called the permittivity of free space. The vector potential produced by a stationary charge is zero.

The potential energy between two stationary charges is equal to the scalar potential produced by one charge times the value of the other charge:
\begin{displaymath} U = \frac{q_1 q_2}{4 \pi \epsilon_0 r} . \end{displaymath} (17.2)

Notice that it doesn't make any difference whether one multiplies the scalar potential from charge 1 by charge 2 or vice versa - the result is the same.

Since $r = (x^2 + y^2 + z^2 )^{1/2}$, the electric field produced by a charge is
\begin{displaymath} \mbox{\bf E} = - \left( \frac{\partial \phi}{\partial x} , ... ...tial z} \right) = \frac{q \mbox{\bf r}}{4 \pi \epsilon_0 r^3} \end{displaymath} (17.3)

where $\mbox{\bf r} = (x,y,z)$ is the vector from the charge to the point where the electric field is being measured. The magnetic field is zero since the vector potential is zero.

The force between two stationary charges separated by a distance $r$ is obtained by multiplying the electric field produced by one charge by the other charge. Thus the magnitude of the force is
\begin{displaymath} F = \frac {q_1 q_2}{4 \pi \epsilon_0 r^2} ~~~ \mbox{(Coulomb's law)}, \end{displaymath} (17.4)

with the force being repulsive if the charges are of the same sign, and attractive if the signs are opposite. This is called Coulomb's law.

Equation (16.4) is the electric equivalent of Newton's universal law of gravitation. Replacing mass by charge and $G$ by $-1/(4 \pi \epsilon_0 )$ in the equation for the gravitational force between two point masses gives us equation (16.4). The most important aspect of this result is that both the gravitational and electrostatic forces decrease as the square of the distance between the particles."

PLEASE SEE THE SITES BELOW FOR LEGIBLE MATH.

The problem is elementary at this point, my dear watson.

Answer 2

Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge

Answer 3

The electric field \mathbf{E} at a given point is defined as the (vectorial) force \mathbf{F} that would be exerted on a stationary test particle of unit charge by electromagnetic forces (i.e. the Lorentz force). A particle of charge q would be subject to a force \mathbf{F}=q.\mathbf{E}.

Its SI units are newtons per coulomb (N⋅C−1) or, equivalently, volts per metre (V⋅m−1), which in terms of SI base units are kg⋅m⋅s−3⋅A−

Answer 4

Electric field E= - dV /dr As potential is constant dV / dr =zero Hence electric field is zero in a region where electric potential is constant. For example, inside a hollow charged matallic sphere, THE ELECTRIC FIELD IS ZERO BUT THE POTENTIAL INSIDE IS CONSTANT and equal to potential on the suface of charged sphere

Answer 5

Electric fields are one of my strongest points beings that no other living sole on Earth has solved this besides me!! ...Try this one..Hydro-Magnetic Energy...Or C\C that's Constant current in a Relative both small or Large Space depending on Design application. A energy form that needs no fuel or Wind or solar power or water flow to self Manipulate it's Relative Motion in it's time and space....